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#### • Class 11 Physics Demo

Explore Related Concepts

#### • in a binomial distribution the random variable x is continuous

From Wikipedia

Continuous probability distribution

In probability theory, a continuous probability distribution is a probability distribution which possesses a probability density function. Mathematicians also call such distribution absolutely continuous, since its cumulative distribution function is absolutely continuous with respect to the Lebesgue measureÎ». If the distribution of X is continuous, then X is called a continuous random variable. There are many examples of continuous probability distributions: normal, uniform, chi-squared, and others.

Intuitively, a continuous random variable is the one which can take a continuous range of values â€” as opposed to a discrete distribution, where the set of possible values for the random variable is at most countable. While for a discrete distribution an event with probability zero is impossible (e.g. rolling 3Â½ on a standard die is impossible, and has probability zero), this is not so in the case of a continuous random variable. For example, if one measures the width of an oak leaf, the result of 3Â½ cm is possible, however it has probability zero because there are infinitely many other potential values even between 3 cm and 4 cm. Each of these individual outcomes has probability zero, yet the probability that the outcome will fall into the interval is nonzero. This apparent paradox is resolved by the fact that the probability that X attains some value within an infinite set, such as an interval, cannot be found by naively adding the probabilities for individual values. Formally, each value has an infinitesimally small probability, which statistically is equivalent to zero.

Formally, if X is a continuous random variable, then it has a probability density functionÆ’(x), and therefore its probability to fall into a given interval, say is given by the integral

\Pr[a\le X\le b] = \int_a^b f(x) \, dx In particular, the probability for X to take any single value a (that is ) is zero, because an integral with coinciding upper and lower limits is always equal to zero.

The definition states that a continuous probability distribution must possess a density, or equivalently, its cumulative distribution function be absolutely continuous. This requirement is stronger than simple continuity of the cdf, and there is a special class of distributions, singular distributions, which are neither continuous nor discrete nor their mixture. An example is given by theCantor distribution. Such singular distributions however are never encountered in practice.

Note on terminology: some authors use the term to denote the distribution with continuous cdf. Thus, their definition includes both the (absolutely) continuous and singular distributions.

Random variable

In probability and statistics, a random variable or stochastic variable is a variable whose value is not known. Its possible values might represent the possible outcomes of a yet-to-be-performed experiment, or the potential values of a quantity whose already-existing value is uncertain (e.g., as a result of incomplete information or imprecise measurements). Intuitively, a random variable can be thought of as a quantity whose value is not fixed, but which can take on different values; a probability distribution is used to describe the probabilities of different values occurring. Realizations of a random variable are called random variates.

Random variables are usually real-valued, but one can consider arbitrary types such as boolean values, complex numbers, vectors, matrices, sequences, trees, sets, shapes, manifolds, functions, and processes. The term random elementis used to encompass all such related concepts. A related concept is thestochastic process, a set of indexed random variables (typically indexed by time or space).

## Introduction

Real-valued random variables (those whose range is the real numbers) are used in the sciences to make predictions based on data obtained from scientific experiments. In addition to scientific applications, random variables were developed for the analysis of games of chance and stochastic events. In such instances, the function that maps the outcome to a real number is often the identity function or similarly trivial function, and not explicitly described. In many cases, however, it is useful to consider random variables that are functions of other random variables, and then the mapping function included in the definition of a random variable becomes important. As an example, the square of a random variable distributed according to a standard normal distribution is itself a random variable, with a chi-square distribution. One way to think of this is to imagine generating a large number of samples from a standard normal distribution, squaring each one, and plotting a histogram of the values observed. With enough samples, the graph of the histogram will approximate the density function of a chi-square distribution with one degree of freedom.

Another example is the sample mean, which is the average of a number of samples. When these samples are independent observations of the same random event they can be called independent identically distributed random variables. Since each sample is a random variable, the sample mean is a function of random variables and hence a random variable itself, whose distribution can be computed and properties determined.

One of the reasons that real-valued random variables are so commonly considered is that the expected value (a type of average) and variance (a measure of the "spread", or extent to which the values are dispersed) of the variable can be computed.

There are two types of random variables: discrete and continuous. A discrete random variable maps outcomes to values of a countable set (e.g., the integers), with each value in the range having probability greater than or equal to zero. A continuous random variable maps outcomes to values of an uncountable set (e.g., the real numbers). For a continuous random variable, the probability of any specific value is zero, whereas the probability of some infinite set of values (such as an interval of non-zero length) may be positive. A random variable can be "mixed", with part of its probability spread out over an interval like a typical continuous variable, and part of it concentrated on particular values like a discrete variable. These classifications are equivalent to the categorization of probability distributions.

The expected value of random vectors, random matrices, and similar aggregates of fixed structure is defined as the aggregation of the expected value computed over each individual element. The concept of "variance of a random vector" is normally expressed through a covariance matrix. No generally-agreed-upon definition of expected value or variance exists for cases other than just discussed.

## Examples

The possible outcomes for one coin toss can be described by the state space \Omega = {heads, tails}. We can introduce a real-valued random variable Y as follows:

Y(\omega) = \begin{cases} 1, & \text{if} \ \ \omega = \text{heads} ,\\ 0, & \text{if} \ \ \omega = \text{tails} . \end{cases}

If the coin is equally likely to land on either side then it has a probability mass function given by:

\rho_Y(y) = \begin{cases}\frac{1}{2},& \text{if }y=1,\\

\frac{1}{2},& \text{if }y=0.\end{cases}

Question:Two dice are rolled and the two numbers on the uppermost faces are added together. What is the probability that this total is (a) six, (b) more than six (c) this dice rolling is carried out 10 times. What is the probability that a total of more than six will occur on exactly 4 occasions. I'm up to learning how to approximate the binomial distribution to a normal distribution, and this question just really confuses me. It's very different from the examples in my book and every time I worked out the answer, it's way off from the real actual answer. My answer for part (a) is 1/6, but apparently it's wrong. I have no idea what i need to do for these questions, so could someone please explain it to me? Thanks Stanschim: Erm..is there a quick and easier way to do part (a)?

Answers:Lets set up the distribution; there are 36 possible outcomes, each equally likely. Die1 Die2 Total 1 1 2 1 2 3 1 3 4 1 4 5 1 5 6 1 6 7 2 1 3 2 2 4 2 3 5 2 4 6 2 5 7 2 6 8 3 1 4 3 2 5 3 3 6 3 4 7 3 5 8 3 6 9 4 1 5 4 2 6 4 3 7 4 4 8 4 5 9 4 6 10 5 1 6 5 2 7 5 3 8 5 4 9 5 5 10 5 6 11 6 1 7 6 2 8 6 3 9 6 4 10 6 5 11 6 6 12 a) There are 5 ways to get 6, so the probability is 5/36. b) There are 21 ways to get more than 6, so the probability is 21/36 = 7/12. c) We just showed that the probability of getting more than 6 is 7/12. We can use that with the binomial distribution to answer your question. (10 nCr 4)(7/12)^(4)(5/12)^6 = 0.1272386405 There is no substitute for "long and boring." Probability is about counting possible events (the denominator) and counting the desired outcomes (the numerator). So, I would say there is no shortcut - boring doesn't take as long as you think!!!

Question:A) # of petals on a randomly chosen daisy. B) Stem length in cm of a random daisy C) # of daisys found in a randomly chosen grassy area 1 sq meter in size. D) average number of petals per daisy computed from all the daisies found in a randomly chosen grassy area 1 sq meter in size. I think A) is discrete and B) is continuous, but what about the rest and why> Thank you.

Answers:you were correct with A and B. C is discrete ( you can't have a fraction of a daisy - only a whole number) D is continuous ( the average could be any number)

Question:Please help..I trying to help my grandson You just can give the steps A continuous random variable X that can assume values only between X=2 and X=8 inclusive has a density function given by f(x) = 1/48(x+3) Show that it is complete p.d.f also Find E(X).

Answers:You must show that the integral of the density function is 1 The integral of 1/48(x + 3) = 1/48 (1/2 x^2 + 3x) Evaluated from 2 to 8 = 1/48 ((1/2(8^2) + 3 (8)) - (1/2(2^2) + 3(2))) = 1 E(x) is the integral of x(f(x))dx = 1/48 (1/3 (x^3) + 1/2 (x^2)) Evaluated from 2 to 8 1/48 ((1/3(8^3) + 1/2(8^2)) - (1/3(2^3) + 1/2(2^2))) = 99/24 (double check, I'm working by hand.)

Question:Is a discrete random variable difined by a probability density function? And is as continuous random variable defined by just a probability distribution? Thanks Vikram!

Answers:A discrete random variable is one which may take on only a countable number of distinct values such as 0,1,2,3,4,........ Discrete random variables are usually (but not necessarily) counts. If a random variable can take only a finite number of distinct values, then it must be discrete. Examples of discrete random variables include the number of children in a family, the Friday night attendance at a cinema, the number of patients in a doctor's surgery, the number of defective light bulbs in a box of ten. A continuous random variable is one which takes an infinite number of possible values. Continuous random variables are usually measurements. Examples include height, weight, the amount of sugar in an orange, the time required to run a mile. If you want more detailed answer then visit http://en.wikipedia.org/wiki/Random_variable

Example on finding the median and quartiles of a continuous random variable. :In this video I run through a numerical example on finding the median and lower and upper quartiles of a continuous random variable from its probability density function. To see this video in a larger clearer form goto ExamSolutions at www.examsolutions.co.uk

Example on finding the Mean E(X) and Variance Var(X) for a Continuous Random Variable :In this example you are shown how to calculate the mean, E(X) and the variance Var(X) for a continuous random variable. To see this video in a larger clearer form goto ExamSolutions at www.examsolutions.co.uk


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# Practice on Toph

Participate in exhilarating programming contests, solve unique algorithm and data structure challenges and be a part of an awesome community.

# Homework

By subeen · Limits 1s, 512 MB

Matthew has purchased a new game and isn't feeling like completing his home work his teacher gave it to him. Can you help Matthew to complete his home work ?

The homework is :
Given an integer $n$, find the minimum number of digits which needs to be removed from $n$ so that $n$ becomes a perfect square without leading zeros. If there is no possible solution print $-1$.

## Input

First line contains an integer $t$ $(1 \leq t \leq 50)$, the number of test cases.

Following $t$ lines, contain an integer $n$ $(1 \leq n \leq 10^{10})$.

## Output

Print $t$ lines with answer for each $n$.

## Sample

InputOutput
3
6434
625
333
2
0
-1

### Statistics

56% Solution Ratio

Lazy_ProgrammerEarliest, Nov '18

Tareq_AbrarFastest, 0.0s

LegendHabibLightest, 0 B

jahir.raihanShortest, 565B

format: plain


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As of May 4 2007 the scripts will autodetect your timezone settings. Nothing here has to be changed, but there are a few things

## Sunday, August 28, 2011

### Contined fractions (2a)

I found a better way to present the table which shows the algorithm for calculating continued fractions:

 $k$ $a_k$ $p_k$ $q_k$ $C_k$ $-1$ $0$ 1 $0$ $1$ $0$ $k$ $a_k$ $a_k \cdot p_{k-1} + p_{k-2}$ $a_k \cdot q_{k-1} + q_{k-2}$ $\frac{p_k}{q_k}$

The table consists of $m+2$ rows. The value of the FCF is $\frac{p_m}{q_m}$.

To be continued.

- Continued fractions (1)
- Continued fractions (2)

## Welcome to The Bridge

Mathematics: is it the fabric of MEST?
This is my voyage
My continuous mission
To uncover hidden structures
To create new theorems and proofs
To boldly go where no man has gone before

(Raumpatrouille – Die phantastischen Abenteuer des Raumschiffes Orion, colloquially aka Raumpatrouille Orion was the first German science fiction television series. Its seven episodes were broadcast by ARD beginning September 17, 1966. The series has since acquired cult status in Germany. Broadcast six years before Star Trek first aired in West Germany (in 1972), it became a huge success.)


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ma104-mid-F08 - MA104 Midterm Test Page 1 of 2 1 Evaluate the following integrals[5 marks(a sin2 x dx cos3 x[5 marks(b tan3(4x dx[5 marks(c cos5 x sin4

# ma104-mid-F08 - MA104 Midterm Test Page 1 of 2 1 Evaluate...

• Notes
• 2

This preview shows page 1 - 2 out of 2 pages.

MA104 — Midterm Test October 23, 2008 Page 1 of 2 1. Evaluate the following integrals.(a)sin2xcos3xdx[5 marks](b)tan3(4x)dx[5 marks](c)cos5xsin4x dx[5 marks]2. By using a suitable trigonometric substitution, evaluate the integral63x2-9x2dx.[8 marks]3. (a) Evaluate the improper integralπ/20sin 2xcosxdxor show that it is divergent.[5 marks](b) Use the Comparison Theorem to determine whether the improper integralπsin2xxdx[5 marks]is convergent or divergent.4. LetRbe the region bounded by the curvesy= 1-x2andy= 4-4x2. Find the volume of[10 marks] 0 5. Find the arc length of the curve defined byy=x-x2+ sin-1(x), 0x1. [ 8 marks ] 6. Find an equation of the curvey=f(x) that satisfies the differential equationdydx=xcosx2y+e3y[9 marks]


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# Damped and undamped oscillations (integration?)

## Homework Statement

Hi everybody! I'm doing a problem about oscillations, and I must admit that a few things are still unclear to me about that subject. Can someone maybe help me?

a) A onedimensional masspoint m is oscillating under the influence of the force $$F(x) = -c⋅x (c > 0).$$ What is the Newton's equation of motion for the system and integrate it with an exponential approach $$x(t) ∼ e^{λt}.$$ Find the integration constant for the initial conditions $$x(t = 0) = x_0$$ and $$\dot{x}(t = 0) = \dot{x_0}.$$
b) A friction force $$-ϒ \dot{x} (ϒ > 0)$$ is added to the system of a), that means a total force $$F(x,\dot{x}) = -cx - ϒ\dot{x}$$ is acting upon the masspoint m. Similarly to question a), find the path $$x(t)$$ and distinguish it between the cases:
$$0 < \frac{ϒ}{m} < 2ω_0 : \mbox{underdamping}$$
$$\frac{ϒ}{m} = 2ω_0 : \mbox{critical damping}$$
$$0 < 2ω_0 < \frac{ϒ}{m} : \mbox{overdamping}$$

## The Attempt at a Solution

a) So first I set up my equation of motion:

$$m \cdot \ddot{x} = -c \cdot x ⇔ m \cdot \ddot{x} + c \cdot x = 0 ⇔ \ddot{x} + \frac{c}{m} x = 0 \\ \mbox{Let } ω_0 = \sqrt{\frac{c}{m}} ⇒ \boldsymbol{\ddot{x} + ω_0^2 x = 0}$$
Then I rewrite the equation in the exponential form and solve for λ:
$$x(t) = α \cdot e^{λt} ⇒ α \cdot λ^2 \cdot e^{λt} + α \cdot ω_0^2 \cdot e^{λt} = 0 ⇔ λ^2 + ω_0^2 = 0 \\ ⇒ \boldsymbol{λ = ± i \cdot ω_0} ⇒ x(t) = α \cdot e^{i ω_0 t} + α^{*} \cdot e^{-i ω_0 t}$$
Then do I have to integrate the equation of motion by substituting x(t) by that expression? Why would we want to do that anyway? To get the velocity? It's the first time I'm ever asked to integrate an equation of motion, and there is no explanation whatsoever with the problem about why should one do that. Anyway I gave it a go:

$$\int -α \cdot ω_0^2 \cdot e^{i ω_0 t} + α \cdot ω_0^2 \cdot e^{-i ω_0 t} dt = -α \cdot ω_0^2 \int e^{i ω_0 t} dt+ α \cdot ω_0^2 \int e^{-i ω_0 t} dt \\ = i \cdot α \cdot ω_0 (e^{i ω_0 t} + \frac{1}{e^{-i ω_0 t}}) + C$$

Ehem...Not sure what is that, if it's correct and what to do from there! I do not know what α is, and all those imaginary units kind of scare me! Some help/hint here would be very appreciated. :)

Julien.

BvU
Homework Helper
Hi Julien,
Then do I have to integrate the equation of motion
You've done that by solving ! You assumed a solution of the given form and found it satisfies the differential equation if ##\lambda = i\omega## and also if ##\lambda = -i\omega##. Since the equation is linear, ##x(t) = A e^{i\omega t} + B e^{-i\omega t}## also satisfies the equation of motion

 I think you already imposed "x(t) is real" to get a restriction the integration constatns.
You restricted yourself with ##a## and ##a^*## if you mean ##a^*## is the complex conjucate of ##a##.

JulienB
@BvU Aaah then no integration needed! That question really confused me

Okay so then if I resume from there:

$$x(t) = Ae^{iω_0t} + Be^{-iω_0t} ⇒ \dot{x}(t) = iω_0Ae^{iω_0t} - iω_0Be^{-iω_0t}$$

The initial conditions can then be expressed in terms of A and B:

$$x(t = 0) = x_0 = A + B \\ \dot{x}(t = 0) = \dot{x}_0 = iω_0A - iω_0B$$

When I solve for A and B I get:

$$A = x_0 - B ⇒ \dot{x_0} = x_0iω_0 - 2Biω_0 ⇒ B = \frac{x_0}{2} - \frac{\dot{x_0}}{2iω_0} ⇒ A = \frac{x_0}{2} + \frac{\dot{x_0}}{2iω_0}$$

Does that make sense? I won't pretend I perfectly understand what I'm doing, so I also don't really know if I have to go further or if that was the question.

Julien.

BvU
Homework Helper
You've answered a) in full, but perhaps it becomes more eye-pleasing if you rewrite a little bit using ##e^{i\phi} = \cos\phi + i\sin\phi##: you get something like ##x_0\cos(\omega t) + {\dot x_0\over \omega} (\sin\omega t)## (Do check me here !)

Now part b) will reveal why this approach as in part a) is so useful.

The harmonic oscillator is a very relevant item for a lot of Physics with a capital P, from solid state phics, quantuum mechanics all the way up to the very front of field theory. But also in acoustics, mechanics, etc, etc. You name it. reason: the F(x) = - kx is the simplest desciption (first order approach) of any equilibrium.

So by all means carry on and ask if stuck.

JulienB
@BvU Thanks a lot for your help! I will try to solve b) on my own and will post my progress again, probably tomorrow.

Julien.

@BvU (Btw your expression with cos and sin was correct )

I started b) and got that:

$$m \cdot \ddot{x} = -ϒ \cdot \dot{x} - c \cdot x ⇔ \ddot{x} + \frac{ϒ}{m} \dot{x} + \frac{c}{m} x = 0 \\ \mbox{Let } 2δ = \frac{ϒ}{m} ⇒ \ddot{x} + 2δ\dot{x} + ω_0^2x = 0 \\ x(t) = α \cdot e^{λt} ⇒ λ^2\cdot α \cdot e^{λt} + 2δ \cdot λ \cdot e^{λt} + ω_0^2 \cdot α \cdot e^{λt} = 0 \\ ⇔ λ^2 + 2δλ + ω_0^2 = 0 ⇒ λ = -δ ± \sqrt{δ^2 - ω_0^2}$$

Okay that's something. :) Now I try to go further for the case "underdamped":

$$0 < 2δ < 2ω_0 : \\ \mbox{Let } ω^2 = ω_0^2 - δ^2 ⇒ λ = -ϒ ± \sqrt{-ω^2} = -ϒ ± iω \\ ⇒ x(t) = A \cdot e^{-δt + iωt} + B \cdot e^{-δt - iωt} = e^{-δt}(A \cdot e^{iωt} + B \cdot e^{-iωt})$$

Then I am afraid I encounter something I do not understand again. Although that looks a lot like in a), in my book it says that there is only one solution A and they write x(t) like that:

$$x(t) = A \cdot e^{-ϒt} cos(ωt + φ)$$

Did I do something wrong? How can I go from the last expression to that one?


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# nLab permutation representation

Contents

### Context

#### Representation theory

representation theory

geometric representation theory

# Contents

## Definition

For $G$ a group (tyically a finite group), consider a G-set $(S, \rho)$, hence a set $S$ (typically a finite set), equipped with an action of $G$

$\rho \;\colon\; G \times S \longrightarrow S \,.$

Equivalently this is a group homomorphism

$\rho \;\colon\; G \longrightarrow Aut_{Set}(S)$

from $G$ to the group of permutations of elements of $S$. As such it is a representation of $G$ “by permutations”.

Specifically, if $S$ is a finite set and an isomorphism $S \simeq \{1, 2, 3, \cdots, n\}$ is understood, it is equivalently a group homomorphism

$\rho \;\colon\; G \longrightarrow S_n$

to the symmetric group $S_n$ on $n$ elements.

For $k$ any field (or, more generally, any commutative ring, but one mostly considers fields) this $G$-action may be linearized to a $k$-linear representation of $G$ in an evident way:

###### Definition

(linear permutation representation)

The linear permutation representation of a G-set $(S,\rho)$ is the following $k$-linear representation of $G$:

1. The underlying $k$-vector space is the freely spanned vector space $k[S]$, whose elements (vectors) are the formal linear combinations

$k[S] \;=\; \left\{ v =\underset{ s \in S_{fin} \subset S }{\sum} v_s \, s \;\vert\; S_{fin} \, \text{finite subset} ,\; v_s \in k \right\}$

of elements of $S$ with coefficients in $k$, hence is the $k$-vector space for which $S$ is a canonical linear basis.

2. The linear $G$-action

$k[\rho] \;\colon\; G \times k[\mathbb{C}] \longrightarrow k[\mathbb{C}]$

is given on linear basis-elements $s \in S \hookrightarrow k[S]$ by $\rho$, which uniquely defines it by linearity to act on a general vector as

$k[\rho]\left(g\right) \;\colon\; v \;\mapsto\; \underset{ s \in S_{fin} \subset S }{\sum} v_s \, \rho(g)(s) \,.$

This concept immediately generalizes to groupoid representations and so forth, see also at infinity-action the section Examples – Discrete group actions on sets.

## Properties

### Functoriality

###### Proposition

(functoriality of linear permutation representations)

The construction of linear permutation representations (Def. ) evidently extends to a functor from the category of G-sets $G Set$ to the category of linear representations $G Rep$

$G Set \overset{ \phantom{AA} k[-] \phantom{AA} }{\longrightarrow} G Rep \,.$

Both of these categories are rig categories with respect to disjoint union and Cartesian product on the left, and direct sum and tensor product of representations on the right.

The functor $k[-]$ is canonically a homomorphism of rig-categories in that in that it is canonically a strong monoidal functor for both “addition” and “multiplication” monoidal structures:

$\big(G Set, \sqcup, \times\big) \overset{ \phantom{AA} k[-] \phantom{AA} }{\longrightarrow} \big(G Rep, \oplus, \otimes \big) \,.$

### Comparison from Burnside- to representation ring

Let $G$ be a finite group and assume all G-sets in the following to be finite sets and all linear representations to be finite dimensional.

Consider

1. the Burnside ring $A(G)$, which is the Grothendieck ring of the rig-category $(G Set, \sqcup, \times)$ of finite G-sets;

2. the representation ring $R(G)$, which is the Grothendieck ring of the rig category $(G Rep, \oplus, \otimes)$ of finite-dimensional linear G-representations.

###### Definition

(permutation representations make ring homomorphism from Burnside ring to representation ring)

Since forming $k$-linear permutation representations (Def. ) is a rig-functor $G Set\overset{k[-]}{\longrightarrow} G Rep$ (Prop. ), under passing to Grothendieck rings it induces a ring homomorphism

$K(k[-]) \;\colon\; K( G Set, \sqcup, \times ) = \; A(G) \overset{\phantom{AA} \beta \phantom{AA}}{\longrightarrow} R(R) \; = K( G Rep, \oplus, \otimes)$

from the Burnside ring of $G$ to its representation ring.

This homomorphism is traditionally denoted $\beta$, as shown.

Its kernel is known as the Brauer relations (e.g. Bartel-Dokchitser 11).

###### Remark

(virtual linear permutation representations)

The image of the comparison morphism $\beta = K(k[-])$ (Def. ) may be called the virtual linear permutation representations.

###### Remark

(virtual permutation representations from equivariant stable cohomotopy into equivariant K-theory)

Under the identitification

1. of the Burnside ring with the equivariant stable cohomotopy of the point

$A(G) \;\simeq\; \mathbb{S}_G(\ast)$

(see there)

2. of the representation ring with the equivariant K-theory of the point

$R(G) \;\simeq\; K_G(\ast)$

(see there)

the ring homomorphism of Def. should be image under forming equivariant cohomology of the point of the initial morphism of E-infinity ring spectra

$\mathbb{S} \longrightarrow KU$

from the sphere spectrum to KU.

Noticing that we may regard stable cohomotopy/the sphere spectrum as being the algebraic K-theory of the “field with one element$\mathbb{F}_1$ (see there)

$\mathbb{S} \simeq K \mathbb{F}_1$

we may regard this as extension of scalars along $\mathbb{F}_1 \to \mathbb{C}$ followed by the comparison map between algebraic and topological K-theory:

(1)$\mathbb{S} \simeq K\mathbb{F}_1 \to K\mathbb{C} \to KU \,.$

graphics grabbed from SS19

(equivariant) cohomologyrepresenting
spectrum
equivariant cohomology
of the point $\ast$
cohomology
of classifying space $B G$
(equivariant)
ordinary cohomology
HZBorel equivariance
$H^\bullet_G(\ast) \simeq H^\bullet(B G, \mathbb{Z})$
(equivariant)
complex K-theory
KUrepresentation ring
$KU_G(\ast) \simeq R_{\mathbb{C}}(G)$
Atiyah-Segal completion theorem
$R(G) \simeq KU_G(\ast) \overset{ \text{compl.} }{\longrightarrow} \widehat {KU_G(\ast)} \simeq KU(B G)$
(equivariant)
complex cobordism cohomology
MU$MU_G(\ast)$completion theorem for complex cobordism cohomology
$MU_G(\ast) \overset{ \text{compl.} }{\longrightarrow} \widehat {MU_G(\ast)} \simeq MU(B G)$
(equivariant)
algebraic K-theory
$K \mathbb{F}_p$representation ring
$(K \mathbb{F}_p)_G(\ast) \simeq R_p(G)$
Rector completion theorem
$R_{\mathbb{F}_p}(G) \simeq K (\mathbb{F}_p)_G(\ast) \overset{ \text{compl.} }{\longrightarrow} \widehat {(K \mathbb{F}_p)_G(\ast)} \!\! \overset{\text{Rector 73}}{\simeq} \!\!\!\!\!\! K \mathbb{F}_p(B G)$
(equivariant)
stable cohomotopy
$K \mathbb{F}_1 \overset{\text{Segal 74}}{\simeq}$ SBurnside ring
$\mathbb{S}_G(\ast) \simeq A(G)$
Segal-Carlsson completion theorem
$A(G) \overset{\text{Segal 71}}{\simeq} \mathbb{S}_G(\ast) \overset{ \text{compl.} }{\longrightarrow} \widehat {\mathbb{S}_G(\ast)} \!\! \overset{\text{Carlsson 84}}{\simeq} \!\!\!\!\!\! \mathbb{S}(B G)$

### Characters

The characters of permutation representations are the Burnside marks of the underlying G-sets:

###### Proposition

(mark homomorphism on cyclic groups agrees with characters of corresponding permutation representations)

For $S \in G Set_{fin}$ a finite G-set, for $k$ any field and $k[S] \in Rep_k(G)$ the corresponding permutation representation, the character $\chi_{k[S]}$ of the permutation representation at any $g \in G$ equals the Burnside marks (Def. ) of $S$ under the cyclic group $\langle g\rangle \subset G$ generated by $g$:

$\chi_{k[S]}\big( g \big) \;=\; \left\vert X^{\langle g \rangle} \right\vert \;\in\; \mathbb{Z} \longrightarrow k \,.$

Hence the mark homomorphism (Def. ) of $G$-sets restricted to cyclic subgroups coincides with the characters of their permutation representations.

This statement immediately generalizes from plain representations to virtual representations, hence to the Burnside ring.

(e.g. tom Dieck 09, (2.15))

###### Proof

By definition of character of a linear representation, we have that

$\chi_{k[S]}(g) = tr_{k[S]}(g)$

is the trace of the linear endomorphism $k[S] \overset{g}{\to} k[S]$ of the given permutation representation.

Now the canonical $k$-linear basis for $k[S]$ is of course the set $S$ itself, and so

\begin{aligned} \chi_{k[S]}(g) & = \underset{ s \in S }{\sum} \left\{ \array{ 1 &\vert& g(s) = s \\ 0 &\vert& \text{otherwise} } \right. \\ & = \left\vert S^g \right\vert \\ & = \left\vert S^{\langle g \rangle} \right\vert \end{aligned}

Here in the first step we spelled out the definition of trace in the canonical basis, and in the second step we observed that the fixed point set of a cyclic group equals that of any one of its generating elements.

## Examples

### Basic examples

###### Example

(regular representation)

For $G$ a group, write, for emphasis, $G_s$ for its underlying set. Let

$\array{ G \times G_s &\overset{ \rho_\ell }{\longrightarrow}& G_s \\ (g,s) &\mapsto& g \cdot s }$

be the canonical action of $G$ on itself, by left multiplication in the group. The corresponding linear permutation representation $(k[G_s], k(\rho_\ell))$ (Def. ) is called the regular representation of $G$.

###### Example

(induced representation of the trivial representation)

Let $G$ be a finite group and $H \overset{\iota}{\hookrightarrow} G$ a subgroup-inclusion. Then the induced representation in Rep(G) of the trivial representation $\mathbf{1} \in Rep(H)$ is the permutation representation $k[G/H]$ of the coset G-set $G/H$:

$\mathrm{ind}_H^G\big( \mathbf{1}\big) \;\simeq\; k[G/H] \,.$

This follows directly as a special case of the general formula for induced representations of finite groups (this Example).

See at induced representation of the trivial representation for more.

### Virtual permutation representations

We discuss here examples of the operation of forming virtual linear permutation representations (Remark ), regarded as the canonical ring homomorphism

$A(G) \overset{ \phantom{AA} \beta \coloneqq K(k[-]) \phantom{AA} }{\longrightarrow} R(G)$

from Def. .

For emphasis, notice that among plain linear representation the linear permutation representations generally form but a tiny sub-class, i.e. generically a linear representation is not a linear permutation representation. But this statement may change radically as we pass to virtual representations:

If the ring homomorphism $\beta$ (Def. ) is surjective function, this means that in fact all virtual linear $G$-representation are virtual linear permutation representations. This is not the case for all groups, but it is the case for large classes of groups! This is the content of Prop. below.

Notice that when this is the case, it means that the representation theory of the given group is, in a precise sense, purely combinatorial, or equivalently, in view of (1), that it is fully determined over the absolute ground field $\mathbb{F}_1$.

###### Proposition

(virtual linear reps from virtual permutation reps)

For ground field $k = \mathbb{Q}$ the rational numbers, the comparison morphism

$A(G) \overset{ \beta = K(k[-]) }{\longrightarrow} R(G)$

from Def. , which sends virtual G-sets to their permutation rep virtual linear G-representations,

1. is surjective for $G$ among one of the following classes of finite groups (not mutually exclusive)

1. binary dihedral groups$\;$ $2 D_{2n}$ for (at least) $2 n \leq 12$

2. the general linear group $GL(2,\mathbb{F}_3)$

2. is not surjective for $G = \mathbb{Z}/3 \times Q_8$ (direct product of cyclic group of order 3 with quaternion group or order 8);

3. is injective precisely for cyclic groups,

4. hence is an isomorphism precisely for cyclic groups.

###### Proof

Isomorphy for the case of cyclic groups is spelled out in tom Dieck 09, Example (4.4.4).

Surjectivity for the case of symmetric groups follows from the theory of Young diagrams (Dress 86, section 3), see also Example below for further pointers.

The proof of surjectivity for p-primary groups is due to Segal 72. (As Segal remarks on his first page, it may also be deduced from Feit 67 (14.3). See also Ritter 72.) The proof is recalled as tom Dieck 79, Theorem 4.4.1.

Surjectivity for binary dihedral groups $2 D_{2n}$ for (at least) $2 n \leq 12$, the binary tetrahedral group, binary octahedral group, binary icosahedral group and the general linear group $GL(2,\mathbb{F}_3)$ is checked by computer experiment in Burton-Sati-Schreiber 18.

The non-surjectivity for $G = \mathbb{Z}/3 \times Q_8$ was remarked in Serre 77, p. 104.

To see that injectivity holds at most for cyclic groups, notice that over $k = \mathbb{Q}$ we have that

1. the number of isomorphism classes of irreducible representations of $G$ equals the number of conjugacy classes of cyclic subgroups;

2. the number of isomorphism classes of indecomposable (transitive) G-sets (i.e. $G$-orbit types) is the number of conjugacy classes of all subgroups.

This means that for $G$ not a cyclic group we have that the free abelian group $A(G))$ has more generators than $R(G)$, so that $\beta$ cannot be injective.

A more general analysis of the cokernel of $\beta$ is due to Berz 94, reviewed and expanded on in Hambleton-Taylor 99. See also Bartel-Dokchitser 14, p. 1.

###### Example

(virtual permutation representations of the group of order 2

Let $G = \mathbb{Z}/2$ be the cyclic group of order 2.

It has two conjugacy classes of subgroups,

1. $H =\mathbb{Z}/2$ the group itself,

2. $H = 1$ the trivial group;

and hence two isomorphism classes of transitive G-sets

1. $(\mathbb{Z}/2)/(\mathbb{Z}/2) = \ast$ the point with the trivial action,

2. $(\mathbb{Z}/2)/1 = \mathbb{Z}/2$ the group itself, with the regular action.

The corresponding linear permutation representations (Def. ) are

1. $k[ (\mathbb{Z}/2)/(\mathbb{Z}/2)] \;\simeq\; \mathbf{1}$,

the 1-dimensional trivial representation;

2. $k[ (\mathbb{Z}/2)/1 ] \; \simeq\; \mathbf{1} \oplus \mathbf{1}_{alt}$,

the direct sum of the 1d trivial representation with the alternating representation.

To see the second item, observe that the non-trivial element $\sigma \in \mathbb{Z}/2$ is represented on $k[\mathbb{Z}/2] \simeq \langle e,\sigma\rangle$ by the permutation matrix

$\left( \array{ 0 & 1 \\ 1 & 0 } \right) \,,$

which is diagonalizable over $k = \mathbb{Z}$ with eigenvectors

1. $\left[\array{ 1 \\ 1 }\right]$ of eigenvalue $1$, spanning the trivial representation $\mathbf{1}$ of dimension 1;

2. $\left[\array{ 1 \\ -1 }\right]$ of eigenvalue $-1$, spanning the alternating representation $\mathbf{1}_{alt}$ of dimension 1.

Hence, the abelian group underlying the representation ring may be identified with the linear span

$R(\mathbb{Z}/2) \;\simeq_k\; \langle \mathbf{1}, \mathbf{1}_{alt} \rangle$

and the comparison morphism from the Burnside ring (Def. ) is

$\array{ A(\mathbb{Z}/2) &\overset{ \phantom{AA} \beta \phantom{AA}}{\longrightarrow}& R(\mathbb{Z}/2) \\ 1\, (\mathbb{Z}/2)/(\mathbb{Z}/2) \;-\; 0\, (\mathbb{Z}/2)/(\mathbb{Z}/2) &\mapsto& \mathbf{1} \\ 1\, (\mathbb{Z}/2)/1 \;-\; 1\, (\mathbb{Z}/2)/(\mathbb{Z}/2) &\mapsto& \mathbf{1}_{alt} \,, }$

which is manifestly an isomorphism, in accord with Prop. .

###### Example

(virtual permutation representations of symmetric groups)

For $G = S_n$ a symmetric group on $n$ elements, the comparison morphism from the Burnside ring to the representation ring (Def. )

$A(S_n) \overset{\beta}{\longrightarrow} R(S_n)$

is a surjective map over $\mathbb{Q}$ but also over $\mathbb{R}$ and $\mathbb{C}$.

The special case of $S_4$ is made explicit for $k =\mathbb{R}$ in Montaldi, bottom of this page, and for $k =\mathbb{C}$ at Categorified Gram-Schmidt process.

## References

Textbook accounts and lecture notes include

Original articles include

• D. L. Johnson, Minimal Permutation Representations of Finite Groups, American Journal of Mathematics Vol. 93, No. 4 (Oct., 1971), pp. 857-866 (jstor:2373739)

• J. Ritter, Ein Induktionssatz fuer rationale Charaktere von nilpotenten Gruppen, J. Reine Angew. Math. 254 (1972), 133–151

• Graeme Segal, Permutation representations of finite $p$-groups, Quart. J. Math. Oxford (2) 23 (1972), 375–381 (doi:10.1093/qmath/23.4.375)

• Jean-Pierre Serre, Linear Representations of Finite Groups, Graduate Texts in Math., vol. 42, Springer–Verlag, New York, 1977

• Andreas Dress, Congruence relations characterizing the representation ring of the symmetric group, Journal of Algebra 101, 350-364 (1986) (pdf, pdf)

• G. Berz, Permutationsbasen fuer endliche Gruppen, Ph.D. thesis, Augsburg, 1994 (Zbl0924.20003)

• I. Hambleton, L. R. Taylor, Rational permutation modules for finite groups, Math. Z. 231 (1999), 707–726 (pdf)

• Alex Bartel, Tim Dokchitser, Brauer relations in finite groups, J. Eur. Math. Soc. 17 (2015), 2473-2512 (arXiv:1103.2047)

• Alex Bartel, Tim Dokchitser, Rational representations and permutation representations of finite groups, Math. Ann. 364 no. 1 (2016), 539-558 (arXiv:1405.6616)

• Vladimir V. Kornyak, An Algorithm to Decompose Permutation Representations of Finite Groups: Polynomial Algebra Approach (arXiv:1801.09786)


text:
# The Nil geometry

## What is Nil?

Nil is a 3-dimensional nilpotent Lie group. It can also be seen as the universal cover of the suspension of a 2-torus by a Dehn twist.

Click on the button below to reveal a concrete model of Nil.

Nil is also the 3-dimensional Heisenberg group, that is the set of matrices of the form $$\begin{bmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{bmatrix}$$ However this model does not highlight the symmetries of Nil. Therefore, we use a different (but isomorphic) model. As a set of points, Nil is the usual 3-dimensional space $X = \mathbb R^3$ with coordinates $(x,y,z)$. The riemanian metric on Nil is given by $$ds^2 = dx^2 + dy^2 + \left(dz - \frac 12(xdy - ydx)\right)^2.$$

Consider to points $p_1 = (x_1, y_1, z_1)$ and $p_2 = (x_2, y_2, z_2)$ in $X$. The group law in this model becomes $$p_1 \ast p_2 = \left(x_1 + x_2, y_1 + y_2, z_1 + z_2 + \frac 12 (x_1 y_2 - x_2 y_1)\right)$$ The left action of Nil on itself is an action by isometries. Its full isometry group is ${\rm Isom}(X) = {\rm Nil} \rtimes O(2)$. In particular the stabilizer of the origin $[0, 0, 0]$ contains a subgroup isomorphic to $S^1$ which corresponds to the rotations around the $z$-axis.

## Some views of Nil

Warning: Some of the real-time simulations below requires a powerful graphic card. If your computer is not fast enough, you can reduce the size of your browser window. Click on the button below to reveal the fly commands.

The default controls to fly in the scene are the following. You can choose your keyboard in the Option controls pannel in the top right corner of the window

Command QWERTY keyboard AZERTY keyboard
Yaw left a q
Yaw right d d
Pitch up w z
Pitch down s s
Roll left q a
Roll right e e
Move forward arrow up arrow up
Move backward arrow down arrow down
Move to the left arrow left arrow left
Move the the right arrow right arrow right
Move upwards ' ù
Move downwards / =
##### Earth in Nil

Earth/Moon system bursting into rings?

##### Tiling

The effect of spiraling geodesics on vision.

##### Lattice of Spheres

The effect of spiraling geodesics on vision.

HD pictures of Nil can be found in the gallery

## Features of Nil

Some features of Nil are described in the following Bridges paper.

##### Geodesics in Nil: Extrinsic

Spray of geodesics in various directions in rotation invariant model.

##### A Sphere In Nil:Extrinsic

The effect of spiraling geodesics on vision.

##### Vertical Planes in Nil

Non-Geodesic Euclidean Planes, with an animated reflecting texture


text:
## DeBroglie

$\lambda=\frac{h}{p}$

405289292
Posts: 52
Joined: Thu Oct 03, 2019 12:16 am

### DeBroglie

I am still sort of confused, what is debroglies equation and what is it describing?

ALegala_2I
Posts: 102
Joined: Thu Jul 11, 2019 12:17 am

### Re: DeBroglie

DeBroglie's proposed that all particles should be considered to have wave like properties. So he devised the equation: wavelength(lambda) = Planck's constant(h)/ momentum(p). In this equation, we are treating light like a wave and finding the wavelength based on the mass and velocity of the particle.

SGonzales_3L
Posts: 55
Joined: Thu Jul 11, 2019 12:17 am

### Re: DeBroglie

Although he derived his equation using light, De Broglie's equation is only valid for particles with mass, so do not use his equation to calculate wavelength for light. His conclusions help us understand that all particles with momentum have wavelike properties that can be calculated using this equation.

Brynne Burrows 3K
Posts: 52
Joined: Thu Jul 25, 2019 12:16 am

### Re: DeBroglie

His equation supports the fact that all matter -- if it is small enough -- has wavelike properties. Using Planck's constant and the momentum (mass*velocity) of the object, the wavelength of that object can be found because it has wavelike properties.


text:
## Set $$X$$ consists of exactly four distinct integers that are greater than $$1.$$ For each integer in the set, all of

##### This topic has expert replies
Legendary Member
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Joined: 07 Sep 2017
Thanked: 6 times
Followed by:5 members

### Set $$X$$ consists of exactly four distinct integers that are greater than $$1.$$ For each integer in the set, all of

by Vincen » Fri Jan 28, 2022 5:04 am

00:00

A

B

C

D

E

## Global Stats

Set $$X$$ consists of exactly four distinct integers that are greater than $$1.$$ For each integer in the set, all of that integer’s unique prime factors are also in the set. For example, if $$20$$ is in set $$X,$$ then $$2$$ and $$5$$ must also be in set $$X.$$ How many of the distinct integers in set $$X$$ are prime?

(1) The product of the integers in set $$X$$ is divisible by $$36.$$

(2) The product of the integers in set $$X$$ is divisible by $$60.$$


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## Problem on computing a determinant by minors and cofactors

Compute the determinant $$\text{det } \begin{pmatrix} 1 & 5 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 3 \end{pmatrix}$$ by minors and cofactors along the second column.
• ## Solution

Let $A$ be the matrix we are seeking the determinant of.
Recall that
Step 1 - We choose to expand across the second column, as directed.
Step 2 - We now form the minors for each entry in the second column.
The minor of the 1,2 entry is the determinant of the submatrix formed by removing the first row and second column: $$\begin{vmatrix} 2 & 0 \\1 & 3\end{vmatrix}.$$
The minor of the 2,2 entry is the determinant of the submatrix formed by removing the second row and the second column: $$\begin{vmatrix} 1 & 0 \\1 & 3\end{vmatrix}.$$
The minor of the 3,2 entry is the determinant of the submatrix formed by removing the third row and the second column: $$\begin{vmatrix} 1 & 0 \\2 & 0\end{vmatrix}.$$
Step 3 - We can write the minor and cofactor expansion along the second column:
\begin{align} \text{det } A = & -5\ \begin{vmatrix} 2& 0 \\ 1 & 3 \end{vmatrix} + 1\ \begin{vmatrix}1 & 0 \\ 1 & 3 \end{vmatrix} - 0\ \begin{vmatrix}1 & 0 \\ 2 & 0 \end{vmatrix} \end{align}
We double check that plus/minus signs agree with the following checkerboard pattern:
The 2,2 position is being added, while the 1,2 and 3,2 positions are being subtracted. Hence, the expansion is correct.
Hence, $$\text{det } A = -5 \cdot 6 + 1 \cdot 3 - 0 \cdot 0 = -27.$$


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## Problem Lotka Volterra Model – Modelling & Plotting in Python

I urgently need your help. Currently I’m conducting a research in regards of the revenue calculation and the dynamics within revenue calculation for my masters. I thought of revenue/profit margin as of a dynamical system – Lotka Volterra differential equations. I thought of a contribution margin calculation as within this simple formula:

In consequence my idea was the following, but I receive an error:

• Can anyone help me?
• Do you think it’s a bad idea to use Lotka Volterra equations for this nonlinear purpose of Sales/Cost/Margin Simulation?
• How would you model it and why? Am I missing equations or does the simple system already fit the requirements?
• How do I generate a valid plot out of these results (Phase Portrait etc.)?

## Plotting a well defined function displays nothing for two-thirds of the range required

The plot in question concerns the second derivative of an inverse Laplace transform (ILT) of a function with five parameters. Here is the ILT

ClearAll["Global*"] prod = (s - cr1) (s - cr2) (s - cr3) (s - cr4); LW = (1 + s)^2/(si prod); Print["symbolic W'=", Wp = D[InverseLaplaceTransform[LW, s, x], x]]

Four parameters are functions of the fifth parameter "si", defined as the roots of a fourth order equation

cr = {cr1, cr2, cr3, cr4} =     s /. Solve[si s^2 + 107 s/5 + 10 ((1 + s)^(-2) - 1) - 1/10 == 0,       s];

Plotting the first derivative of the ILT takes .64

lx = 13; Timing[  pd = Plot[Evaluate[Wp /. si -> 1], {x, 0, lx},     PlotRange -> {{0, lx}, {0.0225, .0275}}]]

Plotting of the second derivative of the ILT takes 14.84 and displays nothing for two-thirds of the range lx

    Wd = D[Wp, x]; Timing[Plot[(Wd /. si -> 1), {x, 0, lx},   PlotRange -> {{0, lx}, {-0.002, .002}}]]

## Recursively plotting specific elements in a list of lists

I am trying to build band structures and am looking to generate a general code that will work regardless of my list size. Here is what I am trying to do: say I have a list like

list = {{0.0, 1, 2, 3}, {0.1, 4, 5, 6}, {0.2, 7, 8, 9}, {0.3, 10, 11, 12}};

and I want to plot the nth element of each list against the first element of each list. How can I tell Mathematica to automatically produce a set of lists to plot like shown below

k = {0.0, 0.1, 0.2, 0.3}; band1 = {1, 4, 7, 10}; band2 = {2, 5, 8, 11}; band3 = {3, 6, 9, 12}; Show[ListLinePlot[k, band1], ListLinePlot[k, band2], ListLinePlot[k, band3]]

Doing it manually one band at a time is possible but a waste of time when I have n=36+ bands. Any help would be much appreciated!

## Plotting data from a list of rules, infer axis titles from the rules with units

I have data, with units, organized as a list of rules; this is so I can slice the data in different ways based on the variables represented.

I’m trying to find the most direct way to plot this and get axis titles with both the variable from the rule and unit from the values. I don’t want to manually label the plots each time, so if Plot doesn’t have a built-in then how can I write a function to do the inference? I’m looking for the most DRY approach possible.

This gives me the units, but not the variable names, in the labels:

data = {    y -> {0, 33.2} (Quantity[1, ("KipsForce")/("Inches")^2]),    x -> {0, 6} 10^-4    }; ListLinePlot[Transpose@({x, y} /. data), AxesLabel -> Automatic]

On the same example, is it possible to retain units like "in/in"? I’ve done this by manually specifying the labels and using HoldForm to avoid evaluating the cancellation, but I can’t get that to work with automatic labels as above.

## Plotting surfaces in 4D

So I am new to Mathematica and I will probably express myself badly, please let me know if I should specify the problem better.

I have the following constraints over the list {a,b,c,d}:

a=f[c,d] b=g[c,d]

So that effectively I have {f[c,d],g[c,d],c,d} and I would like to plot this in a 3D+Color graph, with possibly one of the functions as color. I tried various combinations of Table and ListDensityPlot3D but I wasn’t able to reach anything (apart from a sparsely colored cube).

Moreover, I have a second surface {f2[c,d], g2[c,d],c,d}, is it possible to plot the two together and possibly find intersection points?

Thanks for the help.

## Why are these two histograms plotting so differently?

I have two lists, which are pretty similar. They are given by:

list1=Flatten[Table[#[[1]], #[[2]]] & /@ {{1, 2},{2, 9},{3, 3},{4,73},{5, 38}}] list2=Flatten[Table[#[[1]], #[[2]]] & /@ {{1, 2},{2, 10},{3, 8},{4,65},{5, 22}}]

ie. list1 is a list of 2 1’s, 9 2’s, 3 3’s, etc.

If I plot a histogram of them next to each other they come out like this:

{Histogram[list1],Histogram[list2]}

And I can’t figure out why the labelling is different. I would like to have all of my numbers in the centre of the column, so tried using the answer from this question. as so:

{Histogram[list1,    Ticks -> {{# + .5, #} & /@ HistogramList[list1][[1]], Automatic}],   Histogram[list2,    Ticks -> {{# + .5, #} & /@ HistogramList[list2][[1]], Automatic}]}

But that comes out like this:

Any idea what’s going on?

## Phase portrait plotting by mathematica

I am trying to do the phase space plot of the following expression

$$\frac{dy}{dt} = -0.006y (1-0.001 y^{2})^ {2.003}$$

Anyone can suggest me an appropriate code in mathematica?

## Problem plotting expression involving Generalized hypergeometric functions $_2F_2 \left(.,.,. \right)$

I’m trying to plot a graph for the following expectation

$$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right]=a 2^{-\frac{\kappa }{2}-1} b^{-\frac{\kappa }{2}} \theta ^{-\kappa } \left(\frac{\, _2F_2\left(\frac{\kappa }{2}+\frac{1}{2},\frac{\kappa }{2};\frac{1}{2},\frac{\kappa }{2}+1;\frac{1}{2 b \theta ^2}\right)}{\Gamma \left(\frac{\kappa }{2}+1\right)}-\frac{\kappa \, _2F_2\left(\frac{\kappa }{2}+\frac{1}{2},\frac{\kappa }{2}+1;\frac{3}{2},\frac{\kappa }{2}+\frac{3}{2};\frac{1}{2 b \theta ^2}\right)}{\sqrt{2} \sqrt{b} \theta \Gamma \left(\frac{\kappa +3}{2}\right)}\right)$$ where $$a$$ and $$b$$ are constant values, $$\mathcal{Q}$$ is the Gaussian Q-function, which is defined as $$\mathcal{Q}(x) = \frac{1}{\sqrt{2 \pi}}\int_{x}^{\infty} e^{-u^2/2}du$$ and $$\gamma$$ is a random variable with Gamma distribition, i.e., $$f_{\gamma}(y) \sim \frac{1}{\Gamma(\kappa)\theta^{\kappa}} y^{\kappa-1} e^{-y/\theta}$$ with $$\kappa > 0$$ and $$\theta > 0$$.

This equation was also found with Mathematica, so it seems to be correct. I’ve got the same plotting issue with Matlab.

Follows some examples, where I have checked the analytical results against the simulated ones.

When $$\kappa = 12.85$$, $$\theta = 0.533397$$, $$a=3$$ and $$b = 1/5$$ it returns the correct value $$0.0218116$$.

When $$\kappa = 12.85$$, $$\theta = 0.475391$$, $$a=3$$ and $$b = 1/5$$ it returns the correct value $$0.0408816$$.

When $$\kappa = 12.85$$, $$\theta = 0.423692$$, $$a=3$$ and $$b = 1/5$$ it returns the value $$-1.49831$$, which is negative. However, the correct result should be a value around $$0.0585$$.

When $$\kappa = 12.85$$, $$\theta = 0.336551$$, $$a=3$$ and $$b = 1/5$$ it returns the value $$630902$$. However, the correct result should be a value around $$0.1277$$.

Therefore, the issue happens as $$\theta$$ decreases. For values of $$\theta > 0.423692$$ the analytical matches the simulated results. The issue only happens when $$\theta <= 0.423692$$.

I’d like to know if that is an accuracy issue or if I’m missing something here and if there is a way to correctly plot a graph that matches the simulation. Perhaps there is another way to derive the above equation with other functions or there might be a way to simplify it and get more accurate results.

## Plotting the Eigenvectors with respect to a parameter

I have a matrix of the form given below with a parameter $$\lambda$$. I would like to plot the quantity <$$\phi_{i}|Q|\phi_{i}>$$ for the every Eigenvectors corresponding to ascending order in Eigenvalues of this matrix with respect $$\lambda$$. I am bit trouble to sort the eigenvectors and plot it w.r.t $$\lambda$$. Pl somebody help me to get that. Here {$$\lambda$$,0,1.0,0.01}. Here $$|\phi_{i}>$$ are the eigen vector of the matrix M.

M[\[Lambda]_] =  { {1, 1 + \[Lambda], -2, 5, \[Lambda], 0}, {0, Sqrt[\[Lambda]] + 3, 6, 7, 0, -3}, {1, 4, 6, \[Lambda] + 2, 0, 1}, {0.6, \[Lambda], 6, 4, 8, 0.5}, {Sqrt[2], 3, 11, Sqrt[\[Lambda] + 4], 0, 1}, {4, 0, \[Lambda], 6, 5, 2}}

## Plotting 2 functions at the same time with Manipulate[]

Here is a code that works well:

f1[x_] := Sqrt[25 - x^2] Manipulate[  Plot[f1[x], {x, from, to}, AspectRatio -> Automatic,    PlotRange -> {{-20, 20}, {-20, 20}},    Epilog -> {Text["From y: " <> ToString[f[from]],       Scaled[{1, 1}], {1, 1}],      Text["To y: " <> ToString[f[to]],       Scaled[{1, 0.96}], {1, 1}]}], {{from, -10, "from x"}, -10, 10,    Appearance -> "Labeled"}, {{to, 10, "to x"}, -10, 10,    Appearance -> "Labeled"}]

output like:

Now I want to plot 2 functions at the same time. I tryied something like this:

f1[x1_] := Sqrt[25 - x1^2] f2[x2_] := -Sqrt[25 - x2^2] Manipulate[  Plot[{f1[x1_], f2[x2_]}, {x1, from1, to1}, {x2, from2, to2},    AspectRatio -> Automatic, PlotRange -> {{-20, 20}, {-20, 20}},   Epilog -> {Text["From y1: " <> ToString[f[from1]],       Scaled[{1, 1}], {1, 1}],      Text["To y1: " <> ToString[f[to1]], Scaled[{1, 0.96}], {1, 1}]},   Epilog -> {Text["From y2: " <> ToString[f[from2]],       Scaled[{1, 1}], {1, 1}],      Text["To y2: " <> ToString[f[to2]], Scaled[{1, 0.96}], {1, 1}]}],  {{from1, -10, "from1 x"}, -10, 10,    Appearance -> "Labeled"}, {{to1, 10, "to x"}, -10, 10,    Appearance -> "Labeled"},   {{from2, -10, "from2 x"}, -10, 10,    Appearance -> "Labeled"}, {{to2, 10, "to2 x"}, -10, 10,    Appearance -> "Labeled"}] `

But it does not plots anything. How to correct my code above?


text:
# 17th International Conference on Ion Sources

15-20 October 2017
CERN
Europe/Zurich timezone

## Diagnostics of Ta deposited Plasma Electrode for Negative Hydrogen Ion Production with DC Laser Photodetachment Method

Not scheduled
15m
CERN

#### CERN

Centre international de Conférence Genève (CICG). http://www.cicg.ch/
Poster presentation Negative ion sources

### Speaker

Shingo Masaki (Doshisha University)

### Description

The difference in material that covers the plasma electrode surface can affect the production and destruction processes of negative hydrogen ions (H$^-$). Tantalum adsorption on the plasma electrode is believed to reduce negative ion destruction by capturing hydrogen atoms, and the same effect is expected for caesium adsorption.

In this experiment, tantalum and tungsten filaments produce a hot cathode discharge plasma in a cylindrical ion source filled with hydrogen gas. The filaments evaporate the material to form a layer of each element on the plasma electrode surface. These filaments are 0.35 mm in diameter and the ends are fixed to the current feed through ports with screw to make their length to be about 90 mm. The ion source has 150 mm inner diameter and 200 mm height. A 350 l/s turbo molecular pump coupled to an oil rotary pump evacuate the apparatus to reach the ultimate pressure down to 3.0 × 10$^{-5}$ Pa. Electron temperature and electron density are measured with a Langmuir probe immersed in the plasma. Irradiation of an 808 nm semiconductor laser yields photon induced signal corresponding to the H$^-$ density by photodetachment reaction near the plasma electrode. The characteristics of photodetachment signal against plasma electrode bias are being investigated for different coverage of Ta.

### Primary author

Shingo Masaki (Doshisha University)

### Co-authors

Mamiko Sasao (Doshisha University) Motoi Wada (Doshisha University)


text:
Magma - Maple Programming Help

Home : Support : Online Help : Mathematics : Algebra : Magma : Magma/Rank

Magma

 Rank
 compute the (lexicographic) rank of a magma represented by its Cayley table

 Calling Sequence Rank( m )

Parameters

 m - Array representing the Cayley table of a finite magma

Description

 • The Rank command returns the lexicographic rank of the magma represented by the Cayley table m.

Examples

 > $\mathrm{with}\left(\mathrm{Magma}\right):$
 > $m≔⟨⟨⟨1|2|3⟩,⟨2|3|1⟩,⟨3|1|2⟩⟩⟩$
 ${m}{≔}\left[\begin{array}{ccc}{1}& {2}& {3}\\ {2}& {3}& {1}\\ {3}& {1}& {2}\end{array}\right]$ (1)
 > $\mathrm{Rank}\left(m\right)$
 ${4070}$ (2)
 > $m≔⟨⟨⟨1|2|3⟩,⟨2|3|3⟩,⟨3|1|2⟩⟩⟩$
 ${m}{≔}\left[\begin{array}{ccc}{1}& {2}& {3}\\ {2}& {3}& {3}\\ {3}& {1}& {2}\end{array}\right]$ (3)
 > $\mathrm{Rank}\left(m\right)$
 ${4124}$ (4)

Compatibility

 • The Magma[Rank] command was introduced in Maple 15.


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# Uncertainty on sum of values sampled from unknown probability distribution function

I am dealing with a Monte-Carlo simulation which provides me with a value $x$ as a result. I can run the simulation $n$ times, which gives me $n$ values. I don't know in advance what the probability distribution function of the output values is. What I want to be able to do is to place an uncertainty on the sum or the mean of these values, but I need the uncertainty to be sensible even for small values of $n$.

For example, if I have $n=2$, and the two resulting values of $x$ are very similar (say 9.9 and 10.1), then intuitively, I know that the mean is still not necessarily accurate, because there are only two samples (whereas a standard error in the mean would give me a small uncertainty, because it uses the standard deviation).

Is there a formal way to deal with this kind of situation? In a way, I'm looking for something analogous to Poisson statistics, which allows one to place an uncertainty even on a single value, but in a more generalized way.

-
If you have any guess on the probability distribution of the result, you can either: (1) find the best parameters that describe your distribution, and estimate the uncertainty from there; or (2) use some prior on the probability distribution and proceed Bayesically. The simplest example is assuming that everything is normal, estimating the standard deviation, and concluding the approximate distribution of the mean. –  Yuval Filmus Jul 31 '10 at 23:58
Could you provide more information? As you have described the problem, the simulation could even fail to terminate for some inputs, which makes problems of bounding uncertainty really hard. –  András Salamon Aug 1 '10 at 2:37


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# R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry

R S Aggarwal Solutions for Class 10 Chapter 6 Coordinate Geometry is an outstanding study material formulated for students studying in Class 10. Chapter 6 of class 10 Maths mainly deals with problems based on distance between two points, quadrilateral properties, section formula, mid point and many more. In order to score good marks in the class 10 maths board exam, students must practice R S Aggarwal Solutions for Maths. All solutions are crafted as per the latest CBSE syllabus by subject experts at BYJU’S. Students can avail the R S Aggarwal Solutions and download the pfd for free.

## Download PDF of R S Aggarwal Solutions for Class 10 Chapter 6 Coordinate Geometry

### Access Solutions to Maths R S Aggarwal Chapter 6 – Coordinate Geometry

Get detailed solutions for all the questions listed under the below exercises:

Exercise 6A Solutions: 20 Questions (20 Short Answers)

Exercise 6B Solutions: 20 Questions (20 Short Answers)

Exercise 6C Solutions: 31 Questions (31 Long Answers)

Exercise 6D Solutions: 17 Questions (12 Short Answers & 5 Long Answers )

### Exercise 6A Page No: 307

Question 1: Find the distance between the points:

(i) A (9, 3) and B (15, 11)

(ii) A (7, -4) and B (-5, 1)

(iii) A (-6, -4) and B (9, -12)

(iv) A (1, -3) and B (4, -6)

(v) P (a + b, a – b) and Q (a – b, a + b)

(vi) P (a sin α, a cos α) and Q (a cos α, -a sin α)

Solution:

Question 2: Find the distance of each of the following points from the origin:

(i) A (5, -12)

(ii) B (-5, 5)

(iii) C (-4, -6)

Solution:

Question 3: Find all possible values of x for which the distance between the points A (x, -1) and B (5, 3) is 5 units.

Solution:

Question 4: Find all possible values of y for which the distance between the points A (2, -3) and B (10, y) is 10 units.

Solution:

Question 5: Find the values of x for which the distance between the points P (x, 4) and Q (9, 10) is 10 units.

Solution:

Question 6: If the point A (x, 2) is equidistant from the points B (8, -2) and C (2, -2), find the value of x. Also, find the length of AB.

Solution:

Given: Point A (x, 2) is equidistant from B (8, -2) and C (2, -2)

Which implies:

AB = AC

Squaring both sides

AB^2 = AC^2

Using distance formula:

Question 7: If the point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), find the value of p. Also, find the length of AB.

Solution:

Question 8: Find the point on the x-axis which is equidistant from the points (2, -5) and (-2, 9).

Solution:

Let point P(x, 0) is on x-axis and equidistant from A(2, -5) and B (-2, 9)

PA = PB

or PA^2 = PB^2

(2 – x)^2 + (– 5 – 0)^2 = (– 2 – x)^2 + (9 – 0)^2

(2 – x)^2 + (– 5)^2 = (– 2 – x)^2 + (9)^2

29 + x^2 – 4x = 85 + x^2 + 4x

56 = – 8x

or x = – 7

The point on ×-axis is (– 7, 0)

Question 9: Find points on the x-axis, each of which is at a distance of 10 units from the point A (11, -8).

Solution:

Let the points on x-axis be P(x, 0) and Q(y, 0) which are at distance of 10 units from point A(11, -8).

Which implies:

PA = QA

or PA^2 = QA^2

Either (x – 17) = 0 or (x – 5) = 0

x = 17 or x = 5

So, the points are : (17, 0) and (5, 0)

Question 10: Find the point on the y-axis which is equidistant from the points A (6, 5) and B (-4, 3).

Solution:

Let point P(0, y) is on the y-axis, then

PA = PB

or PA^2 = PB^2

36 = 4y

or y = 9

The required point is (0,9).

Question 11.

If the points P (x, y) is equidistant from the points A (5, 1) and B (-1, 5), prove that 3x = 2y.

Solution:

Since P (x, y) is equidistant from A (5, 1) and B (-1, 5), then

PA = PB

or PA^2 = PB^2

(5 – x)^2 + (1 – y)^2 = (– 1 – x)^2 + (5 – y)^2

(25 + x^2 – 10x) + (1 + y^2 – 2y) = (1 + x^2 + 2x + 25 + y^2 – 10y)

26 + x^2 – 10x + y^2 – 2y = (26 + x^2 + 2x + y^2 – 10y)

12x = 8y

3x = 2y

Hence proved.

Question 12: If P (x, y) is a point equidistant from the points A(6, -1) and B(2, 3), show that x – y = 3.

Solution:

Since P (x, y) is equidistant from A(6, -1) and B(2, 3), then

PA = PB

or PA^2 = PB^2

(6 – x)^2 + (-1 – y)^2 = (2 – x)^2 + (3 – y)^2

(36 + x^2 – 12x) + (1 + y^2 + 2y) = (4 + x^2 – 4x + 9 + y^2 – 6y)

37 – 12x + 2y = 13 – 4x – 6y

8x = 8y + 24

x – y = 3

Hence proved.

Question 13: Find the coordinates of the point equidistant from three given points A (5, 3), B (5, -5) and C (1, -5).

Solution:

Let the coordinates of the point be O(x, y), then

OA = OB = OC

or OA^2 = OB^2 = OC^2

Question 14: If the points A (4, 3) and B (x, 5) lie on a circle with the centre O (2, 3), find the value of x.

Solution:

Given: Points A (4, 3) and B (x, 5) lie on a circle with centre O (2, 3)

To find: value of x

OA = OB

or OA^2 = OB^2

The value of x is 2.

Question 15: If the point C (-2, 3) is equidistant from the points A (3, -1) and B (x, 8), find the values of x. Also, find the distance BC.

Solution:

Given: Point C(-2, 3) is equidistant from points A(3, -1) and B(x,8).

Then

Question 16: If the point P(2, 2) is equidistant from the points A (-2, k) and B(-2k, -3), find k, Also, find the length of AP.

Solution:

Given: Point P(2, 2) is equidistant from the two points A(-2, k) and B(-2k, -3)

Question 17:

(i) If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.

(ii) If the distance of P(x, y) from A(5, 1) and B(-1, 5) are equal then prove that 3x = 2y.

Solution:

(i)

Let point P(x, y) is equidistant from A(a + b, b – a) and B(a – b, a + b), then

AP = BP or AP^2 = BP^2

((a + b) – x)^2 + ((a – b) – y)^2 = ((a – b) – x)^2 + ((a + b) – y)^2

(a + b)^2 + x^2 – 2(a + b)x + (a – b)^2 + y^2 – 2(a – b)y = (a – b)^2 + x^2 – 2(a – b)x + (a + b)^2 + y^2 – 2(a + b)y

(a^2 + b^2 + 2ab + x^2 – 2(a + b)x + b^2 + a^2 – 2ab + y^2 – 2(a – b)y = (a^2 + b^2 – 2ab + x^2 – 2(a – b)x + b^2 + a^2 + 2ab + y^2 – 2(a + b)y

=> – 2(a + b)x – 2(a – b)y = – 2(a – b)x – 2(a + b)y

=> ax + bx + ay – by = ax – bx + ay + by

=> bx = ay

(ii)

Point P(x, y) is equidistant from the points A(5, 1) and B(– 1, 5), means PA = PB or PA^2 = PB^2

(5 – x)^2 + (1 – y)^2 = (– 1 – x)^2 + (5 – y)^2

(25 + x^2 – 10x) + (1 + y^2 – 2y) = (1 + x^2 + 2x + 25 + y^2 – 10y)

26 + x^2 – 10x + y^2 – 2y = (26 + x^2 + 2x + y^2 – 10y)

12x = 8y

3x = 2y

Hence proved.

Question 18: Using the distance formula, show that the given points are collinear:

(i) (1, -1), (5, 2) and (9, 5)

(ii) (6, 9), (0, 1) and (-6, -7)

(iii) (-1, -1), (2, 3) and (8, 11)

(iv) (-2, 5), (0, 1) and (2, -3)

Solution:

Points are collinear if sum of any two of distances is equal to the distance of the third.

Question 19: Show that the points A(7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right triangle.

Solution:

Given points are A(7, 10), B(-2, 5) and C(3, -4)

From above, two of the sides are of equal length, so triangle ABC is an isosceles triangle.

Check for: Isosceles right triangle

Sum of square of two sides = Square of third side

AB^2 + BC^2 = 106 + 106 = 212 = CA^2

Hence given points are vertices of an isosceles right triangle.

Question 20: Show that the points A (3, 0), B (6, 4) and C (-1, 3) are the vertices of an isosceles right triangle.

Solution:

Given points are A (3, 0), B (6, 4) and C (-1, 3)

From above, two of the sides are of equal length, so triangle ABC is an isosceles triangle.

Check for : Isosceles right triangle

Sum of square of two sides = Square of third side

AB^2 + AC^2 = 25 + 25 = 50 = CB^2

Hence given points are vertices of an isosceles right triangle.

## Exercise 6B Page No: 320

Question 1:

(i) Find the coordinates of the point which divides the join of A (-1, 7) and B (4, -3) in the ratio 2 : 3.

(ii) Find the coordinates of the point which divides the join of A (-5, 11) and B (4, -7) in the ratio 7 : 2.

Solution:

If a point P(x,y) divides a line segment having end points coordinates (x_1, y_1) and (x_2, y_2), then coordinates of the point P can be find using below formula:

$x = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$ $y = \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$

(i) Let P(x, y) be the point which divides the line joining the points A (-1, 7) and B (4, -3) in the ratio 2 : 3. then

x = (2 × 4 + 3 × (– 1))/(2 + 3)

= (8 – 3) /5

= 5/5

= 1

x = 1.

y = (2 × – 3 + 3 × 7)/ 5

= (– 6 + 21)/5

= 15 / 5

= 3

y = 1

Therefore, required point is (1, 3).

x = (7 × 4 + 2 × (– 5))/(7 + 2)

= (28 – 10) /9

= 18/9

= 2

y = (7 × (– 7) + 2 × 11)/ 9

= (– 49 + 22)/9

= – 27 / 9

= – 3

Therefore, required point is (2, – 3)

Question 2: Find the coordinates of the points of trisection of the line segment joining the points A (7, -2) and B (1, -5).

Solution:

Let A(7, -2) and B(1, -5) be the given points and P(x, y) and Q (x’, y’) are the points of trisection.

Question 3: If the coordinates of points A and B are (-2, -2) respectively, find the coordinates of the point P such that AP = 3/7 AB, where P lies on the line segment AB.

Solution:

Coordinate of point P(x, y) can be calculated by using below formula:

Now,

$x = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$ $y = \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$

x = ((3 × 2) + 4(– 2))/ (3 + 4)

= (6 – 8)/7

= – 2/7

y = (3(– 4) + 4(– 2))/ 7

= (– 12 – 8)/ 7

= – 20 / 7

Point P is (-2/7, -20/7)

Question 4: Point A lies on the line segment PQ joining P (6, -6) and Q (-4, -1) in such a way that PA/PQ = 2/5. If the point A also lies on the line 3x + k(y + 1) = 0, find the value of k.

Solution:

Let the point A(x, y) which lies on line joining P(6, -6) and Q(-4, -1) such that PA/PQ = 2/5

Line segment PQ is divided by the point A in the ratio 2:3.

$x = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$ $y = \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$

Step 1: Find coordinates of A(x, y)

x = (2(– 4) + 3(6))/(2 + 3)

= (– 8 + 18) /5

= 10/5 = 2

y = (2(– 1) + 3(– 6))/ 5

= (–2 – 18)/5

= – 20 / 5

= – 4

Step 2: Point A also lies on the line 3x + k(y + 1) = 0

3(2) + k(– 4 + 1) = 0

6 – 3k = 0

or k = 2

Question 5: Points P, Q, R and S divide the line segment joining the points A (1, 2) and B (6, 7) in five equal parts. Find the coordinates of the points P, Q and R.

Solution:

Given: Points P, Q, R and S divides a line segment joining the points A (1, 2) and B (6, 7) in 5 equal parts.

We know that:

$x = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$ $y = \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$

Now,

Step 1: Find coordinates of P.

P(x, y) divides AB in the ratio 1:4

x = (1 × 6 + 4 × 1)/1 + 4

= (6 + 4) /5

= 10/5

= 2

y = (1x 7 + 4 × 2)/5

= (7 + 8)/5

= 15 / 5

= 3

So, P(x, y) = P(2, 3)

Step 2: Find coordinates of Q.

Q divides the segment AB in ratio 2:3

x = (2x 6 + 3x 1)/5

= (12 + 3) /5

= 15/5 = 3

y = (2 × 7 + 3 × 2)/ 5

= (14 + 6)/5

= 20 / 5 = 4

So, Q(x, y) = Q(3,4)

Step 3: Find coordinates of R.

R divides the segment AB in ratio 3:2

x = (3 × 6 + 2 × 1)/5

= (18 + 2) /5

= 20/5

= 4

y = (3 × 7 + 2 × 2)/ 5

= (21 + 4)/5

= 25 / 5

= 5

So, R(x, y) = R(4,5)

Question 6: Points P, Q and R in that order are dividing a line segment joining A (1, 6) and B (5, -2) in four equal parts. Find the coordinates of P, Q and R.

Solution:

Given: Points P, Q and R in order divide a line segment joining the points A (1, 6) and B (5, -2) in four equal parts.

Using formulas:

$x = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$ $y = \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$

Step 1: Find coordinates of P.

P(x, y) divides AB in the ratio of 1:3

x = (5+3)/4 = 8/4 = 2

y = (-2+18) = 16/4 = 4

So P(x, y) = P(2, 4)

Step 2: Find coordinates of Q.

Q divides the segment AB in ratio 2:2 or 1:1. So Q ia midpoint of AB

So Q((1+5)/2, (6-2)/2) = (3, 2)

So, Q(x, y) = Q(3,2)

Step 3: Find coordinates of R.

R divides the segment AB in ratio 3:1

x = (3 × 5 + 1 × 1)/4

= 4

y = (3x(-2) + 1×6)/4

= 0

So, R(x, y) = R(4,0)

Question 7: The line segment joining the points A (3, -4) and B (1, 2) is trisected at the points P(p, -2) and Q(1/2, q). Find the values of p and q.

Solution:

The line segment joining the point A(3, -4) and B(1, 2) is trisected by the points P(p, -2) and Q( 1/2, q). (given)

Step 1: Find x coordinate of P which is p

P(p, -2) divides AB in the ratio of 1:2

p = (1+6)/3 = 7/3

Step 2: Find coordinates of Q

Q divides the segment AB in ratio 2:1

x = (2×1 + 1×3)/3

= (2 + 3) /3

= 5/3

y = (2×2 + 1(– 4))/3

= (4 – 4)/3

= 0/ 3

= 0

= q

Therefore, p = 7/3 and q = 0

Question 8: Find the coordinates of the midpoint of the line segment joining

(i) A (3, 0) and B (-5, 4)

(ii) P (-11, -8) and Q (8, -2)

Solution:

(i) Midpoint of the line segment joining A (3, 0) and B (-5, 4):

Midpoint = ((x_1 + x_2)/2 , (y_1+y_2)/2)

= ((3-5)/2, (0+4)/2)

= (-1, 2)

(ii) Midpoint of the line segment joining P (-11, -8) and Q (8, -2)

PQ midpoint = ((-11+8)/2, (-8-2)/2)

= (-3/2, -5)

Question 9: If (2, p) is the midpoint of the line segment joining the points A (6, -5) and B (-2, 11), find the value of p.

Solution:

Given: (2, p) is the mid point of the line segment joining the points A (6, -5), B (-2, 11)

To find: the value of p

p = (-5+11)/2 = 6/2 = 3

Question 10: The midpoint of the line segment joining A (2a, 4) and B (-2, 3b) is C (1, 2a + 1). Find the values of a and b.

Solution:

Mid point of the line segment joining the points A(2a, 4) and B (-2, 3b) is C(1, 2a + 1)

Mid point of AB = ( (2a-2)/2, (4+3b)/2 ) ….(1)

Mid point of AB = (1, 2a + 1) ….(2) (given)

Now, from (1) and (2)

1 = (2a-2)/2

=> a = 2

and 2a + 1 = (4+3b)/2

10-4 = 3b

or b = 2

Answer: a = 2 and b = 2

Question 11: The line segment joining A(-2, 9) and B(6, 3) is a diameter of a circle with centre C. Find the coordinates of C.

Solution:

The line segment joining the points A(-2, 9) and B(6, 3) is a diameter of a circle with centre C.

Which means C is the midpoint of AB.

let (x, y) be the coordinates of C, then

x = (-2 + 6)/2 = 2 and

y = (9+3)/2 = 6

So, coordinates of C are (2, 6).

Question 12: Find the coordinates of a point A, where AB is a diameter of a circle with centre C(2, -3) and the other end of the diameter is B (1, 4).

Solution:

Given:

AB is diameter of a circle with centre C.

Coordinates of C(2, -3) and other point is B (1, 4)

Point C is the midpoint of AB.

Let (x, y) be the coordinates of A, then

2 = (x+1)/2

4 = x +1

x = 3

and

-3 = (y+4)/2

-6 =y +4

or y = -10

So, coordinates of A are (3, -10).

Question 13: In what ratio does the point P(2, 5) divide the join of A(8, 2) and B(-6, 9)?

Solution:

Given: P (2, 5) divides the line segment joining the points A(8,2) and B(-6, 9).

Let P divides the AB in the ratio m : n

Question 14: Find the ratio in which the point P( 3/4, 5/12) divides the line segment joining the points A(1/2, 3/5) and B (2, -5).

Solution:

Let P divides the line segment joining the points A and B in the ratio m:n.

3n – 2n = 8m – 3n

or m/n = 1/5

Therefore, required ratio is 1:5.

Question 15: Find the ratio in which the point P (m, 6) divides the join of A (-4, 3) and B (2, 8). Also, find the value of m.

Solution:

Let P divides the join of A and B in the ratio k : 1, then

Step 1: Find coordinates of P:

6 = (k × 8 + 1×3)/ (k+1)

=> 6k + 6 = 8k + 3

or k = 3/2

P divides the join of A and B in the ratio 3:2

Step 2: Find the value of m

m = (2k-4)/(k+1) = (2×3/2 – 4)/(3/2+1)

= -1/(5/2)

= -2/5

Question 16: Find the ratio in which the point (-3, k) divides the join of A (-5, -4) and B (-2, 3). Also, find the value of k.

Solution:

Let point P divides the join of A and B in the ratio m : n, then

ratio = m:n = 2:1

Now,

The value of k is 2/3.

Question 17: In what ratio is the segment joining A (2, -3) and B (5, 6) divided by the x-axis?

Also, find the coordinates of the point of division.

Solution:

Let point P on the x-axis divides the line segment joining the points A and B the ratio m : n

Consider P lies on x-axis having coordinates (x, 0).

x = (m × 5 + n x2)/ (m + n)

x = (5m + 2n ) / (m + n)

5m + 2n = x(m + n )

(5 – x)m + (2 – x)n = 0 ….(1)

And,

y = 0 = (m × 6 + n(– 3))/ (m+n)

0 = (6m – 3n ) /(m + n)

6m – 3n = 0

6m = 3n

or m/n = 3/6 = 1/2

P divides AB in the ratio 1:2.

=> m = 1 and n = 2

From (2)

(5 – x) + (2 – x)(2) = 0

5 – x + 4 – 2x = 0

3x = 9

x = 3

Hence coordinates are (3,0)

Question 18: In what ratio is the line segment joining the points A (-2, -3) and B (3, 7) divided by the y-axis? Also, find the coordinates of the point of division.

Solution:

Given: A (-2, -3) and B (3, 7) divided by the y-axis

Let P lies on y-axis and dividing the line segment AB in the ratio m : n

So, the coordinates of P be (0, y)

Therefore, the coordinates of P be (0, 1).

Question 19: In what ratio does the line x – y – 2 = 0 divide the line segment joining the points A (3, -1) and B (8, 9)?

Solution:

Given: A line segment joining the points A (3, -1) and B (8, 9) and another line x – y – 2 = 0.

Let a point P (x, y) on the given line x – y – 2 = 0 divides the line segment AB in the ratio m : n

To find: ratio m:n

Since point P lies on x – y – 2 = 0, so

The required ratio is 2:3.

Question 20: Find the lengths of the medians of a ∆ABC whose vertices are A(0, -1), B(2, 1) and C(0, 3).

Solution:

Given: Vertices of ∆ABC are A(0, -1), B(2, 1) and C(0, 3)

Let AD, BE and CF are the medians of sides BC, CA and AB respectively, then

## Exercise 6C Page No: 336

Question 1: Find the area of ∆ABC whose vertices are:

(i) A (1, 2), B (-2, 3) and C (-3, -4)

(ii) A (-5, 7), B (-4, -5) and C (4, 5)

(iii) A (3, 8), B (-4, 2) and C (5, -1)

(iv) A (10, -6), B (2, 5) and C (-1, 3)

Solution:

Area of ∆ABC whose vertices are is (x_1,y_1 ),(x_2,y_2 )and (x_3,y_3 ) are

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

(i) In ∆ABC, vertices are A (1, 2), B (-2, 3) and C (-3, -4)

Area of triangle = 1/2(1(–2 + 3)–2(–4–2)–3(2–3))

= 1/2(1 + 12 + 3)

= 8 sq units

(ii) A (-5, 7), B (-4, -5) and C (4, 5)

Area of triangle = 1/2(–5(–5–5)–4(5–7) + 4(7 + 5))

= 1/2(–50 + 8 + 48)

= 5 sq units

(iii) A (3, 8), B (-4, 2) and C (5, -1)

Area of triangle = 1/2(3(2 + 1)–4(–1–8) + 5(8–2))

= 1/2(9 + 36 + 30)

= 1/2(75)

= 37.5 sq units

(iv) A (10, -6), B (2, 5) and C (-1, 3)

Area of triangle = 1/2(10(5–3) + 2(3 + 6)–1(–6–5))

= 1/2(20 + 18 + 11)

= 1/2(49)

= 24.5 sq units

Question 2: Find the area of quadrilateral ABCD whose vertices are A (3, -1), B (9, -5), C (14, 0) and D (9, 19).

Solution: Vertices of quadrilateral ABCD are A(3,-1), B(9, -5), C (14, 0) and D(9, 19)

Construction: Join diagonal AC.

We know that:

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Question 3: Find the area of quadrilateral PQRS whose vertices are P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2).

Solution:

Given: PQRS is a quadrilateral whose vertices are P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2)

Construction: Join PR

We know that:

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Now, Area of quadrilateral PQRS = Area of triangle PQR + Area of triangle PSR

= 21/2 + 35/2

= 28 sq. units

Question 4: Find the area of quadrilateral ABCD whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4).

Solution:

Given: ABCD is a quadrilateral whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4).

By Joining AC, we get two triangles ABC and ADC

Question 5: If A (-7, 5), B (-6, -7), C (-3, -8) and D (2, 3) are the vertices of a quadrilateral ABCD then find the area of the quadrilateral.

Solution:

Question 6: Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).

Solution:

Given: A triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5)

Let D, E and F are the midpoints of the sides CB, CA and AB respectively of ∆ABC, as shown in the below figure.

Question 7: A (7, -3), B (5, 3) and C (3, -1) are the vertices of a ∆ABC and AD is its median. Prove that the median AD divides ∆ABC into two triangles of equal areas.

Solution:

So,

Area of triangle ABD:

= 1/2(7(3–1) + 5(1+3) + 4(–3–3))

= 1/2(14 + 20–24)

= 1/2(10)

= 5 sq. units …(1)

Area of triangle ACD:

= 1/2(7(–1–1) + 3(1 + 3) + 4(–3 + 1))

= 1/2(–14 + 12–8)

= 1/2(10)

= 5 sq. units ….(2)

From (1) and (2), we conclude that Area of triangle ABD and ACD is equal.

Hence proved.

Question 8: Find the area of ∆ABC with A (1, -4) and midpoints of sides through A being (2, -1) and (0, -1).

Solution:

Given: A ∆ABC with A(1, -4)

Question 9: A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ∆ADE.

Solution:

A(6, 1), B(8, 2) and C(9, 4) are the three vertices of a parallelogram ABCD.

E is the midpoint of DC.

Join AE, AC and BD which intersects at O, where O is midpoint of AC.

Question 10: (i) If the vertices of ∆ABC be A (1, -3), B (4, p) and C (-9, 7) and its area is 15 square units, find the values of p.

(ii) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is

(7/2, y), find the value of y.

Solution:

We know that:

Question 11: Find the value of k so that the area of the triangle with vertices A (k + 1, 1), B (4, -3) and C (7, -k) is 6 square units.

Solution:

Question 12: For what value of k (k > 0) is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) equal to 53 square units?

Solution:

Question 13: Show that the following points are collinear.

(i) A (2, -2), B (-3, 8) and C (-1, 4)

(ii) A (-5, 1), B (5, 5) and C (10, 7)

(iii) A (5, 1), B (1, -1) and C (11, 4)

(iv) A (8, 1), B (3, -4) and C (2, -5)

Solution:

Points are collinear if the area of a triangle is equal to zero.

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

(i) A (2, -2), B (-3, 8) and C (-1, 4)

Δ = 1/2{2 (8– 4) + (–3) (4 + 2) –1 (2– 8)}

Δ = 1/2 {8–18 + 10}

Δ = 0

Hence points are collinear.

(ii) A (-5, 1), B (5, 5) and C (10, 7)

Δ = 1/2{–5(5– 7) + 5 (7–1) + 10 (1–5)}

Δ = 1/2{10 + 30–40}

Δ = 0

Hence points are collinear.

(iii) A (5, 1), B (1, -1) and C (11, 4)

Δ = 1/2{5(–1– 4) + 1 (4– 1) + 11 (1 + 1)}

= 1/2{–25 + 3 + 22}

= 0

Hence points are collinear.

(iv) A (8, 1), B (3, -4) and C (2, -5)

Δ = 1/2{8(–4 + 5) + 3 (–5–1) + 2 (1 + 4)}

= 1/2{8–18 + 10}

= 0

Hence points are collinear.

Question 14: Find the value of x for which the points A (x, 2), B (-3, -4) and C (7, -5) are collinear.

Solution:

Points are A (x, 2), B (-3, -4) and C (7, -5) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(x + 63) = 0

Or x = -63

Question 15: For what value of x are the points A (-3, 12), B (7, 6) and C (x, 9) collinear?

Solution:

Points are A (-3, 12), B (7, 6) and C (x, 9) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(6x -12) = 0

Or x = 2

Question 16: For what value of y are points P (1, 4), Q (3, y) and R (-3, 16) are collinear?

Solution:

Points are P (1, 4), Q (3, y) and R (-3, 16) are collinear.

Which means area of triangle PQR = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(4y + 8) = 0

Or y = -2

Question 17: Find the value of y for which the points A (-3, 9), B (2, y) and C (4, -5) are collinear.

Solution:

Points are A (-3, 9), B (2, y) and C (4, -5) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(-7y – 7) = 0

Or y = -1

Question 18: For what values of k are the points A (8, 1), B (3, -2k) and C (k, -5) collinear.

Solution:

Points are A (8, 1), B (3, -2k) and C (k, -5) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(2k^2 – 15k + 22) = 0

Or 2k^2 – 15k + 22 = 0

2k^2 – 11k – 4k + 22 = 0

k(2k – 11) – 2(2k – 11) = 0

(k-2)(2k -11) = 0

k = 2 or k = 11/2. Answer.

Question 19: Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear.

Solution:

Points are A(2, 1), B(x, y) and C(7, 5) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(4x -5y – 3) = 0

4x – 5y – 3 = 0

Relationship between x and y.

Question 20: Find a relation between x and y, if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear.

Solution:

Points are A(x, y), B(-5, 7) and C(-4, 5) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(2x +y + 3) = 0

2x +y + 3 = 0

Relationship between x and y.

Question 21: Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear, if 1/a + 1/b = 1.

Solution:

Points are A(a, 0), B(0, b) and C(1, 1) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC = \frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

= ½[(a(b-1) + 0(1-0) + 1(0-b)]

= ½[ab – a + 0 – b]

Since points are collinear:

½(ab -a-b) = 0

ab – a – b = 0

Divide each term by “ab”, we get

1 – 1/b -1/a = 0

or 1/a + 1/b = 1. hence proved.

Question 22: If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a + b = 1, find the values of a and b.

Solution:

Points are P(-3, 9), Q(a, b) and R(4, -5) are collinear.

Which means area of triangle = 0

$Area of \bigtriangleup ABC = \frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear, we have

½(-14a – 7b + 21) = 0

-14a – 7b + 21 = 0

2a + b =3 …..(1)

a + b = 1 …..(2) (given)

From (1) and (2)

a = 2 and b = -1

Question 23: Find the area of ∆ABC with vertices A (0, -1), B (2, 1) and C (0, 3). Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 : 1.

Solution:

Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)

Area of triangle DEF:

= 1/2[1+0+1]

= 1 sq. units

Therefore,

Ratio in the area of triangles ABC and DEF = 4/1 = 4:1.

Question 24: If a ≠ b ≠ c, prove that (a, a²), (b, b²), (0, 0) will not be collinear.

Solution:

Let A (a, a²), B (b, b²) and C (0, 0) are the vertices of a triangle.

Let us assume that that points are collinear, then area of ∆ABC must be zero.

Now, area of ∆ABC

Which is contraction to our assumption.

This implies points are not be collinear. Hence proved.

## Exercise 6D Page No: 340

Question 1: Points A(-1, y) and B(5, 7) lie on a circle with centre O(2, -3y). Find the values of y.

Solution:

Points A(-1, y) and B(5, 7) lie on a circle with centre O(2, -3y).

Which means: OA = OB or OA^2 = OB^2

using distance formula, we get

(-1-2)^2 + (y-(-3y))^2 = (5-2)^2 + (7-(-3y))^2

9 + 16y^2 = 9 + (7 + 3y)^2

16y^2 = 49 + 42y + 9y^2

7y^2 – 42y – 49 = 0

7(y^2-6y-7) = 0

y^2-7y + y-7 = 0

y(y-7) + 1(y-7) = 0

(y + 1)(y-7) = 0

Therefore, y = 7 or y = -1

Possible values of y are 7 or -1.

Question 2: If the point A (0, 2) is equidistant from the points B (3, p) and C (p, 5) find p.

Solution:

A (0, 2) is equidistant from the points B (3, p) and C (p, 5)

Which means: AB = AC or AB^2 = AC^2

using distance formula, we get

(0-3)^2 + (2-p)^2 = (0-p)^2 + (2-5)^2

9 + 4 + p^2 – 4p = p^2 + 9

4p-4 = 0

p = 1

Therefore, the value of p is 1.

Question 3: ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3). Find the length of one of its diagonal.

Solution:

ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3).

Find length of one of its diagonal, say BD: using distance formula, we get

Therefore, length of one of its diagonal is 1.

Question 4: If the point P (k – 1, 2) is equidistant from the points A (3, k) and B (k, 5), find the values of k.

Solution:

Point P (k – 1, 2) is equidistant from the points A (3, k) and B (k, 5).

PA = PB or PA^2 = PB^2

k(k-1)-5(k-1) = 0

(k-5)(k-1) = 0

k = 1 or k = 5

Question 5: Find the ratio in which the point P (x, 2) divides the join of A (12, 5) and B (4, -3).

Solution:

If point P (x, 2) divides the join of A (12, 5) and B (4, -3), then

using section formula, we get

2 = (m x (-3) + n x (5)) / (m + n)

2m + 2n = -3m + 5n

5m = 3n

m/n = m:n = 3:5

The required ratio is 3:5.

Question 6: Prove that the diagonals of a rectangle ABCD with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6) are equal and bisect each other.

Solution:

Vertices f a rectangle ABCD are A(2, -1), B(5, -1), C(5, 6) and D(2, 6)

To prove: Diagonals of the rectangle are equal and bisect each other.

Question 7: Find the lengths of the medians AD and BE of ∆ABC whose vertices are A(7, -3), B(5, 3) and C(3, -1).

Solution:

Question 8: If the points C (k, 4) divides the join of A (2, 6) and B (5, 1) in the ratio 2 : 3 then find the value of k.

Solution:

C (k, 4) divides the join of A (2, 6) and B (5, 1) in the ratio 2 : 3.

Using Section Formula:

After simplifying, we get k = 16/5

The value of k is 16/5.

Question 9: Find the point on x-axis which is equidistant from points A (-1, 0) and B (5, 0).

Solution:

Since point lies on x-axis, y-coordinate of the point will be zero.

Let P (x, 0) be on x-axis which is equidistant from A (-1, 0) and B (5, 0)

Using section formula:

Or x = 2

Thus, the required point is (2, 0).

Question 10: Find the distance between the points (-8/2, 2) and (2/5, 2).

Solution:

Using distance formula, we have

Question 11: Find the value of a, so that the point (3, a) lies on the line represented by 2x – 3y = 5.

Solution:

The points (3, a) lies on the line 2x – 3y = 5.

Put value of x = 3 and y = a in given equation,

2 x 3 – 3 x a = 5

6 – 3a = 5

3a = 6 – 5

a = 1/3

Question 12: If the points A (4, 3) and B (x, 5) lie on the circle with centre O(2, 3), find the value of x.

Solution:

Points A (4, 3) and B (x, 5) lie on the circle with centre O(2, 3)

Which means: OA = OB

=> OA^2 = OB^2

2-x = 0

x = 2

The value of x is 2.

Question 13: If P (x, y) is equidistant from the points A (7, 1) and B (3, 5), find the relation between x and y.

Solution:

Question 14: If the centroid of ∆ABC having vertices A (a, b), B (b, c) and C (c, a) is the origin, then find the value of (a + b + c).

Solution:

Centroid of ∆ABC having vertices A (a, b), B (b, c) and C (c, a) is the origin.

Let O (0, 0) is the centroid of ∆ABC.

a + b + c = 0

And

Question 15: Find the centroid of ∆ABC whose vertices are A(2, 2), B(-4, -4) and C(5, -8).

Solution:

Question 16: In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7, 8)?

Solution:

Point C(4, 5) divide the join of A(2, 3) and B(7, 8)

Let point C(4, 5) divides the AB in the ratio m : n

Using section formula:

4m + 4n = 7m + 2n

3m = 2n

m:n = 2:3

The required ratio is 2:3.

Question 17: If the points A(2, 3), B(4, k) and C(6, -3) are collinear, find the value of k.

Solution:

Points A(2, 3), B(4, k) and C(6, -3) are collinear.

Area of triangle having vertices A, B and C = 0

## R S Aggarwal Solutions for Chapter 6 Coordinate Geometry Topics

In this chapter students will study important concepts on Coordinate Geometry as listed below:

• Coordinate Geometry introduction
• Distance between two points.
• Section formula.
• Midpoint Formula.
• Centroid of a triangle.
• Area of a triangle

### Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry

1. R S Aggarwal Solutions for Maths covers all the exercise questions.

2. It helps to accelerate your knowledge on various concepts of Coordinate Geometry.

3. All questions are solved by subject experts.

4. Easy for quick revision.


text:
# A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge.

Question:

A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Solution:

Net electric flux $\left(\Phi_{\text {Net }}\right)$ through the cubic surface is given by,

$\phi_{\text {Net }}=\frac{q}{\epsilon_{0}}$

Where,

$\epsilon_{0}=$ Permittivity of free space

$=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$

$q=$ Net charge contained inside the cube $=2.0 \mu \mathrm{C}=2 \times 10^{-6} \mathrm{C}$

$\therefore \phi_{\text {Net }}=\frac{2 \times 10^{-6}}{8.854 \times 10^{-12}}$

$=2.26 \times 10^{5} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1}$

The net electric flux through the surface is $2.26 \times 10^{5} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1}$.