NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11

NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11 - Straight Lines

Edited By Vishal kumar | Updated on Nov 17, 2023 09:12 AM IST

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 10 : Straight Lines Miscellaneous Exercise- In this chapter, you have already learned about the slope of the line, parallel lines, perpendicular lines, angles between two lines, colinearity of three points, equations of a line in various forms, the distance of a point from a line, etc. In the Class 11 Maths Chapter 10 miscellaneous exercise solutions, you will get mixed equations from all the concepts given in this Class 11 NCERT syllabus chapter. If you are done with previous exercises, you can start solving problems from the miscellaneous exercise.

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Miscellaneous exercise chapter 10 Class 11 is considered to be tougher than the other exercises of this chapter, so you may not able to solve them by yourself at first. NCERT solutions for Class 11 Maths chapter 10 miscellaneous exercise are here to help you to solve them very easily. Questions from the miscellaneous exercises are not much asked in the CBSE exam but these exercises are very important for engineering entrance exams like JEE Main, SRMJEE, etc. The class 11 maths ch 10 miscellaneous exercise solutions are meticulously crafted by highly qualified subject experts from Careers360. These solutions offer detailed and comprehensive explanations, ensuring a thorough understanding of the concepts. Additionally, the availability of a PDF version allows students the flexibility to access the class 11 chapter 10 maths miscellaneous solutions offline, anytime and anywhere, free of charge. This resource is designed to facilitate convenient and effective learning for students studying Straight Lines in Class 11 Mathematics. If you are looking for NCERT Solutions, you can click on the above link where you will NCERT solutions from Classes 6 to 12 at one place.

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**In the CBSE Syllabus for 2023-24, this miscellaneous exercise class 11 chapter 10 has been renumbered as Chapter 9.

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines Miscellaneous Exercise

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Question:1(a) Find the values of $k$ for which the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is

Parallel to the x-axis.

Answer:

Given equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
and equation of x-axis is $y=0$
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of $y=0$ is , $m' = 0$
and
Slope of line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is , $m = \frac{k-3}{4-k^2}$
Now,
$m=m'$
$\frac{k-3}{4-k^2}=0$
$k-3=0$
$k=3$
Therefore, value of k is 3

Question:1(b) Find the values of $k$ for which the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is

Parallel to the y-axis.

Answer:

Given equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
and equation of y-axis is $x = 0$
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of $y=0$ is , $m' = \infty = \frac{1}{0}$
and
Slope of line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is , $m = \frac{k-3}{4-k^2}$
Now,
$m=m'$
$\frac{k-3}{4-k^2}=\frac{1}{0}$
$4-k^2=0$
$k=\pm2$
Therefore, value of k is $\pm2$

Question:1(c) Find the values of k for which the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is Passing through the origin.

Answer:

Given equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
It is given that it passes through origin (0,0)
Therefore,
$(k-3).0-(4-k^2).0+k^2-7k+6=0$
$k^2-7k+6=0$
$k^2-6k-k+6=0$
$(k-6)(k-1)=0$
$k = 6 \ or \ 1$
Therefore, value of k is $6 \ or \ 1$

Question:2 Find the values of $\small \theta$ and $\small p$ , if the equation $\small x\cos \theta +y\sin \theta =p$ is the normal form of the line $\small \sqrt{3}x+y+2=0$ .

Answer:

The normal form of the line is $\small x\cos \theta +y\sin \theta =p$
Given the equation of lines is
$\small \sqrt{3}x+y+2=0$
First, we need to convert it into normal form. So, divide both the sides by $\small \sqrt{(\sqrt3)^2+1^2}= \sqrt{3+1}= \sqrt4=2$
$\small -\frac{\sqrt3\cos \theta}{2}-\frac{y}{2}= 1$
On comparing both
we will get
$\small \cos \theta = -\frac{\sqrt3}{2}, \sin \theta = -\frac{1}{2} \ and \ p = 1$
$\small \theta = \frac{7\pi}{6} \ and \ p =1$
Therefore, the answer is $\small \theta = \frac{7\pi}{6} \ and \ p =1$

Question:3 Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are $\small 1$ and $\small -6$ , respectively.

Answer:

Let the intercepts on x and y-axis are a and b respectively
It is given that
$a+b = 1 \ \ and \ \ a.b = -6$
$a= 1-b$
$\Rightarrow b.(1-b)=-6$
$\Rightarrow b-b^2=-6$
$\Rightarrow b^2-b-6=0$
$\Rightarrow b^2-3b+2b-6=0$
$\Rightarrow (b+2)(b-3)=0$
$\Rightarrow b = -2 \ and \ 3$
Now, when $b=-2\Rightarrow a=3$
and when $b=3\Rightarrow a=-2$
We know that the intercept form of the line is
$\frac{x}{a}+\frac{y}{b}=1$

Case (i) when a = 3 and b = -2
$\frac{x}{3}+\frac{y}{-2}=1$
$\Rightarrow 2x-3y=6$

Case (ii) when a = -2 and b = 3
$\frac{x}{-2}+\frac{y}{3}=1$
$\Rightarrow -3x+2y=6$
Therefore, equations of lines are $2x-3y=6 \ and \ -3x+2y=6$

Question:4 What are the points on the $\small y$ -axis whose distance from the line $\small \frac{x}{3}+\frac{y}{4}=1$ is $\small 4$ units.

Answer:

Given the equation of the line is
$\small \frac{x}{3}+\frac{y}{4}=1$
we can rewrite it as
$4x+3y=12$
Let's take point on y-axis is $(0,y)$
It is given that the distance of the point $(0,y)$ from line $4x+3y=12$ is 4 units
Now,
$d= \left | \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}} \right |$
In this problem $A = 4 , B=3 , C =-12 ,d=4\ \ and \ \ (x_1,y_1) = (0,y)$
$4 = \left | \frac{4\times 0+3\times y-12}{\sqrt{4^2+3^2}} \right |=\left | \frac{3y-12}{\sqrt{16+9}} \right |=\left | \frac{3y-12}{5} \right |$

Case (i)

$4 = \frac{3y-12}{5}$
$20=3y-12$
$y = \frac{32}{3}$
Therefore, the point is $\left ( 0,\frac{32}{3} \right )$ -(i)

Case (ii)

$4=-\left ( \frac{3y-12}{5} \right )$
$20=-3y+12$
$y = -\frac{8}{3}$
Therefore, the point is $\left ( 0,-\frac{8}{3} \right )$ -(ii)
Therefore, points on the $\small y$ -axis whose distance from the line $\small \frac{x}{3}+\frac{y}{4}=1$ is $\small 4$ units are $\left ( 0,\frac{32}{3} \right )$ and $\left ( 0,-\frac{8}{3} \right )$

Question:5 Find perpendicular distance from the origin to the line joining the points $(\cos \theta ,\sin \theta )$ and $(\cos \phi ,\sin \phi ).$ .

Answer:

Equation of line passing through the points $(\cos \theta ,\sin \theta )$ and $(\cos \phi ,\sin \phi )$ is
$(y-\sin \theta )= \frac{\sin \phi -\sin \theta}{\cos \phi -\cos \theta}(x-\cos\theta)$
$\Rightarrow (\cos \phi -\cos \theta)(y-\sin \theta )= (\sin \phi -\sin \theta)(x-\cos\theta)$
$\Rightarrow y(\cos \phi -\cos \theta)-\sin \theta(\cos \phi -\cos \theta)=x (\sin \phi -\sin \theta)-\cos\theta(\sin \phi -\sin \theta)$
$\Rightarrow x (\sin \phi -\sin \theta)-y(\cos \phi -\cos \theta)=\cos\theta(\sin \phi -\sin \theta)-\sin \theta(\cos \phi -\cos \theta)$ $\Rightarrow x (\sin \phi -\sin \theta)-y(\cos \phi -\cos \theta)=\sin(\theta-\phi)$
$(\because \cos a\sin b -\sin a\cos b = \sin(a-b) )$
Now, distance from origin(0,0) is
$d = \left | \frac{(\sin\phi -\sin\theta).0-(\cos\phi-\cos\theta).0-\sin(\theta-\phi)}{\sqrt{(\sin\phi-\sin\theta)^2+(\cos\phi-\cos\theta)^2}} \right |$
$d = \left | \frac{-\sin(\theta-\phi)}{\sqrt{(\sin^2\phi+\cos^2\phi)+(\sin^2\theta+\cos^2\theta)-2(\cos\theta\cos\phi+\sin\theta\sin\phi)}} \right |$
$d = \left | \frac{-\sin(\theta-\phi)}{1+1-2\cos(\theta-\phi)} \right |$ $(\because \cos a\cos b +\sin a\sin b = \cos(a-b) \ \ and \ \ \sin^2a+\cos^2a=1)$
$d = \left |\frac{ - \sin(\theta-\phi)}{2(1-\cos(\theta-\phi))} \right |$
$d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{\sqrt{2.(2\\sin^2\frac{\theta-\phi}{2})}}\right |$
$d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{2\sin\frac{\theta-\phi}{2}}\right |$
$d = \left | \cos\frac{\theta-\phi}{2} \right |$

Question:6 Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines $\small x-7y+5=0$ and $\small 3x+y=0$ .

Answer:

Point of intersection of the lines $\small x-7y+5=0$ and $\small 3x+y=0$
$\left ( -\frac{5}{22},\frac{15}{22} \right )$
It is given that this line is parallel to y - axis i.e. $x=0$ which means their slopes are equal
Slope of $x=0$ is , $m' = \infty = \frac{1}{0}$
Let the Slope of line passing through point $\left ( -\frac{5}{22},\frac{15}{22} \right )$ is m
Then,
$m=m'= \frac{1}{0}$
Now, equation of line passing through point $\left ( -\frac{5}{22},\frac{15}{22} \right )$ and with slope $\frac{1}{0}$ is
$(y-\frac{15}{22})= \frac{1}{0}(x+\frac{5}{22})$
$x = -\frac{5}{22}$
Therefore, equation of line is $x = -\frac{5}{22}$

Question:7 Find the equation of a line drawn perpendicular to the line $\small \frac{x}{4}+\frac{y}{6}=1$ through the point, where it meets the $\small y$ -axis.

Answer:

given equation of line is
$\small \frac{x}{4}+\frac{y}{6}=1$
we can rewrite it as
$3x+2y=12$
Slope of line $3x+2y=12$ , $m' = -\frac{3}{2}$
Let the Slope of perpendicular line is m
$m = -\frac{1}{m'}= \frac{2}{3}$
Now, the ponit of intersection of $3x+2y=12$ and $x =0$ is $(0,6)$
Equation of line passing through point $(0,6)$ and with slope $\frac{2}{3}$ is
$(y-6)= \frac{2}{3}(x-0)$
$3(y-6)= 2x$
$2x-3y+18=0$
Therefore, equation of line is $2x-3y+18=0$

Question:8 Find the area of the triangle formed by the lines $\small y-x=0,x+y=0$ and $\small x-k=0$ .

Answer:

Given equations of lines are
$y-x=0 \ \ \ \ \ \ \ \ \ \ \ -(i)$
$x+y=0 \ \ \ \ \ \ \ \ \ \ \ -(ii)$
$x-k=0 \ \ \ \ \ \ \ \ \ \ \ -(iii)$
The point if intersection of (i) and (ii) is (0,0)
The point if intersection of (ii) and (iii) is (k,-k)
The point if intersection of (i) and (iii) is (k,k)
Therefore, the vertices of triangle formed by three lines are $(0,0), (k,-k) \ and \ (k,k)$
Now, we know that area of triangle whose vertices are $(x_1,y_1),(x_2,y_2) \ and \ (x_3,y_3)$ is
$A = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
$A= \frac{1}{2}|0(-k-k)+k(k-0)+k(0+k)|$
$A= \frac{1}{2}|k^2+k^2|$
$A= \frac{1}{2}|2k^2|$
$A= k^2$
Therefore, area of triangle is $k^2 \ square \ units$

Question:9 Find the value of $\small p$ so that the three lines $\small 3x+y-2=0,px+2y-3=0$ and $\small 2x-y-3=0$ may intersect at one point.

Answer:

Point of intersection of lines $\small 3x+y-2=0$ and $\small 2x-y-3=0$ is $(1,-1)$
Now, $(1,-1)$ must satisfy equation $px+2y-3=0$
Therefore,
$p(1)+2(-1)-3=0$
$p-2-3=0$
$p=5$
Therefore, the value of p is $5$

Question:10 If three lines whose equations are $y=m_1x+c_1,y=m_2x+c_2$ and $y=m_3x+c_3$ are concurrent, then show that $m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0$ .

Answer:

Concurrent lines means they all intersect at the same point
Now, given equation of lines are
$y=m_1x+c_1 \ \ \ \ \ \ \ \ \ \ \ -(i)$
$y=m_2x+c_2 \ \ \ \ \ \ \ \ \ \ \ -(ii)$
$y=m_3x+c_3 \ \ \ \ \ \ \ \ \ \ \ -(iii)$
Point of intersection of equation (i) and (ii) $\left ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} \right )$

Now, lines are concurrent which means point $\left ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} \right )$ also satisfy equation (iii)
Therefore,

$\frac{m_1c_2-m_2c_1}{m_1-m_2}=m_3.\left ( \frac{c_2-c_1}{m_1-m_2} \right )+c_3$

$m_1c_2-m_2c_1= m_3(c_2-c_1)+c_3(m_1-m_2)$

$m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0$

Hence proved

Question:11 Find the equation of the lines through the point $\small (3,2)$ which make an angle of $\small 45^{\circ}$ with the line $\small x-2y=3$ .

Answer:

Given the equation of the line is
$\small x-2y=3$
The slope of line $\small x-2y=3$ , $m_2= \frac{1}{2}$
Let the slope of the other line is, $m_1=m$
Now, it is given that both the lines make an angle $\small 45^{\circ}$ with each other
Therefore,
$\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |$
$\tan 45\degree = \left | \frac{\frac{1}{2}-m}{1+\frac{m}{2}} \right |$
$1= \left | \frac{1-2m}{2+m} \right |$
Now,

Case (i)
$1=\frac{1-2m}{2+m}$
$2+m=1-2m$
$m = -\frac{1}{3}$
Equation of line passing through the point $\small (3,2)$ and with slope $-\frac{1}{3}$
$(y-2)=-\frac{1}{3}(x-3)$
$3(y-2)=-1(x-3)$
$x+3y=9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

Case (ii)
$1=-\left ( \frac{1-2m}{2+m} \right )$
$2+m=-(1-2m)$
$m= 3$
Equation of line passing through the point $\small (3,2)$ and with slope 3 is
$(y-2)=3(x-3)$
$3x-y=7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$

Therefore, equations of lines are $3x-y=7$ and $x+3y=9$

Question:12 Find the equation of the line passing through the point of intersection of the lines $4x+7y-3=0$ and $2x-3y+1=0$ that has equal intercepts on the axes.

Answer:

Point of intersection of the lines $4x+7y-3=0$ and $2x-3y+1=0$ is $\left ( \frac{1}{13},\frac{5}{13} \right )$
We know that the intercept form of the line is
$\frac{x}{a}+\frac{y}{b}= 1$
It is given that line make equal intercepts on x and y axis
Therefore,
a = b
Now, the equation reduces to
$x+y = a$ -(i)
It passes through point $\left ( \frac{1}{13},\frac{5}{13} \right )$
Therefore,
$a = \frac{1}{13}+\frac{5}{13}= \frac{6}{13}$
Put the value of a in equation (i)
we will get
$13x+13y=6$
Therefore, equation of line is $13x+13y=6$

Question:13 Show that the equation of the line passing through the origin and making an angle $\small \theta$ with the line $\small y=mx+c$ is $\small \frac{y}{x}=\frac{m\pm \tan \theta }{1\mp m\tan \theta }$ .

Answer:

Slope of line $\small y=mx+c$ is m
Let the slope of other line is m'
It is given that both the line makes an angle $\small \theta$ with each other
Therefore,
$\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |$
$\tan \theta = \left | \frac{m-m'}{1+mm'} \right |$
$\mp(1+mm')\tan \theta =(m-m')$
$\mp\tan \theta +m'(\mp m\tan\theta+1)= m$
$m'= \frac{m\pm \tan \theta}{1\mp m\tan \theta}$
Now, equation of line passing through origin (0,0) and with slope $\frac{m\pm \tan \theta}{1\mp m\tan \theta}$ is
$(y-0)=\frac{m\pm \tan \theta}{1\mp m\tan \theta}(x-0)$
$\frac{y}{x}=\frac{m\pm \tan \theta}{1\mp m\tan \theta}$
Hence proved

Question:14 In what ratio, the line joining $\small (-1,1)$ and $\small (5,7)$ is divided by the line $x+y=4$ ?

Answer:

Equation of line joining $\small (-1,1)$ and $\small (5,7)$ is
$(y-1)= \frac{7-1}{5+1}(x+1)$
$\Rightarrow (y-1)= \frac{6}{6}(x+1)$
$\Rightarrow (y-1)= 1(x+1)$
$\Rightarrow x-y+2=0$
Now, point of intersection of lines $x+y=4$ and $x-y+2=0$ is $(1,3)$
Now, let's suppose point $(1,3)$ divides the line segment joining $\small (-1,1)$ and $\small (5,7)$ in $1:k$
Then,
$(1,3)= \left ( \frac{k(-1)+1(5)}{k+1},\frac{k(1)+1(7)}{k+1} \right )$
$1=\frac{-k+5}{k+1} \ \ and \ \ 3 = \frac{k+7}{k+1}$
$\Rightarrow k =2$
Therefore, the line joining $\small (-1,1)$ and $\small (5,7)$ is divided by the line $x+y=4$ in ratio $1:2$

Question:15 Find the distance of the line $\small 4x+7y+5=0$ from the point $\small (1,2)$ along the line $\small 2x-y=0$ .

Answer:

22229
point $\small (1,2)$ lies on line $2x-y =0$
Now, point of intersection of lines $2x-y =0$ and $\small 4x+7y+5=0$ is $\left ( -\frac{5}{18},-\frac{5}{9} \right )$
Now, we know that the distance between two point is given by
$d = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|$
$d = |\sqrt{(1+\frac{5}{18})^2+(2+\frac{5}{9})^2}|$
$d = |\sqrt{(\frac{23}{18})^2+(\frac{23}{9})^2}|$
$d = \left | \sqrt{\frac{529}{324}+\frac{529}{81}} \right |$
$d = \left | \sqrt{\frac{529+2116}{324}} \right | = \left | \sqrt\frac{2645}{324} \right | =\frac{23\sqrt5}{18}$
Therefore, the distance of the line $\small 4x+7y+5=0$ from the point $\small (1,2)$ along the line $\small 2x-y=0$ is $\frac{23\sqrt5}{18} \ units$

Question:16 Find the direction in which a straight line must be drawn through the point $\small (-1,2)$ so that its point of intersection with the line $\small x+y=4$ may be at a distance of $3$ units from this point.

Answer:

Let $(x_1,y_1)$ be the point of intersection
it lies on line $\small x+y=4$
Therefore,
$x_1+y_1=4 \\ x_1=4-y_1\ \ \ \ \ \ \ \ \ \ \ -(i)$
Distance of point $(x_1,y_1)$ from $\small (-1,2)$ is 3
Therefore,
$3= |\sqrt{(x_1+1)^2+(y_1-2)^2}|$
Square both the sides and put value from equation (i)
$9= (5-y_1)^2+(y_1-2)^2\\ 9=y_1^2+25-10y_1+y_1^2+4-4y_1\\ 2y_1^2-14y_1+20=0\\ y_1^2-7y_1+10=0\\ y_1^2-5y_1-2y_1+10=0\\ (y_1-2)(y_1-5)=0\\ y_1=2 \ or \ y_1 = 5$
When $y_1 = 2 \Rightarrow x_1 = 2$ point is $(2,2)$
and
When $y_1 = 5 \Rightarrow x_1 = -1$ point is $(-1,5)$
Now, slope of line joining point $(2,2)$ and $\small (-1,2)$ is
$m = \frac{2-2}{-1-2}=0$
Therefore, line is parallel to x-axis -(i)

or
slope of line joining point $(-1,5)$ and $\small (-1,2)$
$m = \frac{5-2}{-1+2}=\infty$
Therefore, line is parallel to y-axis -(ii)

Therefore, line is parallel to x -axis or parallel to y-axis

Question:17 The hypotenuse of a right angled triangle has its ends at the points $\small (1,3)$ and $\small (-4,1)$ Find an equation of the legs (perpendicular sides) of the triangle.

Answer:

22234
Slope of line OA and OB are negative times inverse of each other
Slope of line OA is , $m=\frac{3-y}{1-x}\Rightarrow (3-y)=m(1-x)$
Slope of line OB is , $-\frac{1}{m}= \frac{1-y}{-4-x}\Rightarrow (x+4)=m(1-y)$
Now,

Now, for a given value of m we get these equations
If $m = \infty$
$1-x=0 \ \ \ \ and \ \ \ \ \ 1-y =0$
$x=1 \ \ \ \ and \ \ \ \ \ y =1$

Question:18 Find the image of the point $\small (3,8)$ with respect to the line $x+3y=7$ assuming the line to be a plane mirror.

Answer:

Ch. 10
Let point $(a,b)$ is the image of point $\small (3,8)$ w.r.t. to line $x+3y=7$
line $x+3y=7$ is perpendicular bisector of line joining points $\small (3,8)$ and $(a,b)$
Slope of line $x+3y=7$ , $m' = -\frac{1}{3}$
Slope of line joining points $\small (3,8)$ and $(a,b)$ is , $m = \frac{8-b}{3-a}$
Now,
$m = -\frac{1}{m'} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because lines \ are \ perpendicular)$
$\frac{8-b}{3-a}= 3$
$8-b=9-3a$
$3a-b=1 \ \ \ \ \ \ \ \ \ \ \ -(i)$
Point of intersection is the midpoint of line joining points $\small (3,8)$ and $(a,b)$
Therefore,
Point of intersection is $\left ( \frac{3+a}{2},\frac{b+8}{2} \right )$
Point $\left ( \frac{3+a}{2},\frac{b+8}{2} \right )$ also satisfy the line $x+3y=7$
Therefore,
$\frac{3+a}{2}+3.\frac{b+8}{2}=7$
$a+3b=-13 \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$(a,b) = (-1,-4)$
Therefore, the image of the point $\small (3,8)$ with respect to the line $x+3y=7$ is $(-1,-4)$

Question:19 If the lines $\small y=3x+1$ and $\small 2y=x+3$ are equally inclined to the line $\small y=mx+4$ , find the value of $m$ .

Answer:

Given equation of lines are
$\small y=3x+1 \ \ \ \ \ \ \ \ \ \ -(i)$
$\small 2y=x+3 \ \ \ \ \ \ \ \ \ \ -(ii)$
$\small y=mx+4 \ \ \ \ \ \ \ \ \ \ -(iii)$
Now, it is given that line (i) and (ii) are equally inclined to the line (iii)
Slope of line $\small y=3x+1$ is , $\small m_1=3$
Slope of line $\small 2y=x+3$ is , $\small m_2= \frac{1}{2}$
Slope of line $\small y=mx+4$ is , $\small m_3=m$
Now, we know that
$\tan \theta = \left | \frac{m_1-m_2}{1+m_1m_2} \right |$
Now,
$\tan \theta_1 = \left | \frac{3-m}{1+3m} \right |$ and $\tan \theta_2 = \left | \frac{\frac{1}{2}-m}{1+\frac{m}{2}} \right |$

It is given that $\tan \theta_1=\tan \theta_2$
Therefore,
$\left | \frac{3-m}{1+3m} \right |= \left | \frac{1-2m}{2+m} \right |$
$\frac{3-m}{1+3m}= \pm\left ( \frac{1-2m}{2+m} \right )$
Now, if $\frac{3-m}{1+3m}= \left ( \frac{1-2m}{2+m} \right )$
Then,
$(2+m)(3-m)=(1-2m)(1+3m)$
$6+m-m^2=1+m-6m^2$
$5m^2=-5$
$m= \sqrt{-1}$
Which is not possible
Now, if $\frac{3-m}{1+3m}= -\left ( \frac{1-2m}{2+m} \right )$
Then,
$(2+m)(3-m)=-(1-2m)(1+3m)$
$6+m-m^2=-1-m+6m^2$
$7m^2-2m-7=0$
$m = \frac{-(-2)\pm \sqrt{(-2)^2-4\times 7\times (-7)}}{2\times 7}= \frac{2\pm \sqrt{200}}{14}= \frac{1\pm5\sqrt2}{7}$

Therefore, the value of m is $\frac{1\pm5\sqrt2}{7}$

Question:20 If the sum of the perpendicular distances of a variable point $\small P(x,y)$ from the lines $\small x+y-5=0$ and $\small 3x-2y+7=0$ is always $\small 10$ . Show that $\small P$ must move on a line.

Answer:

Given the equation of line are
$x+y-5=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
$3x-2y+7=0 \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Now, perpendicular distances of a variable point $\small P(x,y)$ from the lines are

$d_1=\left | \frac{1.x+1.y-5}{\sqrt{1^2+1^2}} \right |$ $d_2=\left | \frac{3.x-2.y+7}{\sqrt{3^2+2^2}} \right |$
$d_1=\left | \frac{x+y-5}{\sqrt2} \right |$ $d_2=\left | \frac{3x-2y+7}{\sqrt{13}} \right |$
Now, it is given that
$d_1+d_2= 10$
Therefore,
$\frac{x+y-5}{\sqrt2}+\frac{3x-2y+7}{\sqrt{13}}=10$
$(assuming \ x+y-5 > 0 \ and \ 3x-2y+7 >0)$
$(x+y-5)\sqrt{13}+(3x-2y+7)\sqrt2=10\sqrt{26}$

$x(\sqrt{13}+3\sqrt{2})+y(\sqrt{13}-2\sqrt{2})=10\sqrt{26}+5\sqrt{13}-7\sqrt2$

Which is the equation of the line
Hence proved

Question:21 Find equation of the line which is equidistant from parallel lines $9x+6y-7=0$ and $3x+2y+6=0$ .

Answer:

Let's take the point $p(a,b)$ which is equidistance from parallel lines $9x+6y-7=0$ and $3x+2y+6=0$
Therefore,
$d_1= \left | \frac{9.a+6.b-7}{\sqrt{9^2+6^2}} \right |$ $d_2= \left | \frac{3.a+2.b+6}{\sqrt{3^2+2^2}} \right |$
$d_1= \left | \frac{9a+6b-7}{\sqrt{117}} \right |$ $d_2= \left | \frac{3a+2b+6}{\sqrt{13}} \right |$
It is that $d_1=d_2$
Therefore,
$\left | \frac{9a+6b-7}{3\sqrt{13}} \right |= \left | \frac{3a+2b+6}{\sqrt{13}} \right |$
$(9a+6b-7)=\pm 3(3a+2b+6)$
Now, case (i)
$(9a+6b-7)= 3(3a+2b+6)$
$25=0$
Therefore, this case is not possible

Case (ii)
$(9a+6b-7)= -3(3a+2b+6)$
$18a+12b+11=0$

Therefore, the required equation of the line is $18a+12b+11=0$

Question:22 A ray of light passing through the point $(1,2)$ reflects on the $x$ -axis at point $A$ and the reflected ray passes through the point $(5,3)$ . Find the coordinates of $A$ .

Answer:

22246
From the figure above we can say that
The slope of line AC $(m)= \tan \theta$
Therefore,
$\tan \theta = \frac{3-0}{5-a} = \frac{3}{5-a} \ \ \ \ \ \ \ \ \ \ (i)$
Similarly,
The slope of line AB $(m') = \tan(180\degree-\theta)$
Therefore,
$\tan(180\degree-\theta) = \frac{2-0}{1-a}$
$-\tan\theta= \frac{2}{1-a}$
$\tan\theta= \frac{2}{a-1} \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Now, from equation (i) and (ii) we will get
$\frac{3}{5-a} = \frac{2}{a-1}$
$\Rightarrow 3(a-1)= 2(5-a)$
$\Rightarrow 3a-3= 10-2a$
$\Rightarrow 5a=13$
$\Rightarrow a=\frac{13}{5}$
Therefore, the coordinates of $A$ . is $\left ( \frac{13}{5},0 \right )$

Question:23 Prove that the product of the lengths of the perpendiculars drawn from the points $\small (\sqrt{a^2-b^2},0)$ and $\small (-\sqrt{a^2-b^2},0)$ to the line $\small \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ is $\small b^2$ .

Answer:

Given equation id line is
$\small \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$
We can rewrite it as
$xb\cos \theta +ya\sin \theta =ab$
Now, the distance of the line $xb\cos \theta +ya\sin \theta =ab$ from the point $\small (\sqrt{a^2-b^2},0)$ is given by
$d_1=\left | \frac{b\cos\theta.\sqrt{a^2-b^2}+a\sin \theta.0-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right | = \left | \frac{b\cos\theta.\sqrt{a^2-b^2}-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |$
Similarly,
The distance of the line $xb\cos \theta +ya\sin \theta =ab$ from the point $\small (-\sqrt{a^2-b^2},0)$ is given by
$d_2=\left | \frac{b\cos\theta.(-\sqrt{a^2-b^2})+a\sin \theta.0-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right | = \left | \frac{-(b\cos\theta.\sqrt{a^2-b^2}+ab)}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |$
$d_1.d_2 = \left | \frac{b\cos\theta.\sqrt{a^2-b^2}-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |.\times\left | \frac{-(b\cos\theta.\sqrt{a^2-b^2}+ab)}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |$
$=\left | \frac{-((b\cos\theta.\sqrt{a^2-b^2})^2-(ab)^2)}{(b\cos\theta)^2+(a\sin\theta)^2} \right |$
$=\left | \frac{-b^2\cos^2\theta.(a^2-b^2)+a^2b^2)}{(b\cos\theta)^2+(a\sin\theta)^2} \right |$
$=\left | \frac{-a^2b^2\cos^2\theta+b^4\cos^2\theta+a^2b^2)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |$
$=\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |$
$=\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2(\sin^2\theta+\cos^2\theta))}{b^2\cos^2\theta+a^2\sin^2\theta} \right | \ \ \ \ (\because \sin^2a+\cos^2a=1)$
$=\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2\sin^2\theta-a^2\cos^2\theta)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |$
$=\left | \frac{+b^2(b^2\cos^2\theta+a^2\sin^2\theta)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |$
$=b^2$
Hence proved

Question:24 A person standing at the junction (crossing) of two straight paths represented by the equations $\small 2x-3y+4=0$ and $\small 3x+4y-5=0$ wants to reach the path whose equation is $\small 6x-7y+8=0$ in the least time. Find equation of the path that he should follow.

Answer:

point of intersection of lines $\small 2x-3y+4=0$ and $\small 3x+4y-5=0$ (junction) is $\left ( -\frac{1}{17},\frac{22}{17} \right )$
Now, person reaches to path $\small 6x-7y+8=0$ in least time when it follow the path perpendicular to it
Now,
Slope of line $\small 6x-7y+8=0$ is , $m'=\frac{6}{7}$
let the slope of line perpendicular to it is , m
Then,
$m= -\frac{1}{m}= -\frac{7}{6}$
Now, equation of line passing through point $\left ( -\frac{1}{17},\frac{22}{17} \right )$ and with slope $-\frac{7}{6}$ is
$\left ( y-\frac{22}{17} \right )= -\frac{7}{6}\left ( x-(-\frac{1}{17}) \right )$
$\Rightarrow 6(17y-22)=-7(17x+1)$
$\Rightarrow 102y-132=-119x-7$
$\Rightarrow 119x+102y=125$

Therefore, the required equation of line is $119x+102y=125$

Topic Covered in class 11 Maths ch 10 Miscellaneous Exercise Solutions

The Miscellaneous Exercise in NCERT Solutions for Class 11 Maths Chapter 10 - Straight Lines covers the following topics:

Slope of a Line
Various Forms of the Equation of a Line
General Equation of a Line
Distance of a Point From a Line

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Mathematics, being a subject that benefits from consistent practice, requires a thorough understanding of the best problem-solving methods. By practising with the help of these NCERT Solutions for Class 11 Maths, students can enhance their problem-solving skills, ultimately leading to the potential to score high marks in examinations.

Benefits of NCERT Solutions for Class 11 Maths Chapter 10 Miscellaneous Exercise:-

Class 11 Maths chapter 10 miscellaneous exercise solutions are designed by subject matter experts based on the guideline given by CBSE.
Class 11 Maths chapter 10 miscellaneous solutions are solved in a step-by-step manner that could be understood by an average student also.

Key Features of Class 11 Chapter 10 Maths Miscellaneous Solutions

Comprehensive Coverage: Solutions cover all themiscellaneous exercise class 11 chapter 10, addressing various aspects of Straight Lines.
Step-by-Step Solutions: Each class 11 maths miscellaneous exercise chapter 10 problem is solved with a step-by-step approach, facilitating a clear and logical progression of concepts for Class 11 students.
Clarity in Language: The class 11 chapter 10 miscellaneous exercise solutions are presented in clear and concise language, making complex geometric concepts associated with Straight Lines more accessible for students.
PDF Version Access: The class 11 maths ch 10 miscellaneous exercise solutions are available in a downloadable PDF format, providing students with offline access for study anytime and anywhere.

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Frequently Asked Question (FAQs)

1. Write the equation of line passing through (5,4) and parallel to y-axis ?

Equation of line parallel to y-axis => x = c

Line passing through the point (5,4) => 5 = c

Equation of line => x = 5

2. Write the equation of line passing through (5,4) and parallel to x-axis ?

Equation of line parallel to x-axis => y = c

Line passing through the point (5,4) => 4 = c

Equation of line => y = 4

3. Write the equation of line passing through (1,0) and make angle of 45 degree with positive x-axis ?

Line make angle of 45 degree with positive x-axis => slope of line = 1

Equation of line with slope 1 => y = x + c

Line pass through the (1,0) => 0 = 1 + c => c=-1

Equation of line => y = x -1

4. Find the slope of line 4y = 3x - 6 ?

Given line 4y = 3x - 6

y = 3x/4 - 6/4

Compare with y = mx + c

Slope (m) = 3/4

5. Do I need to buy NCERT solutions book for Class 11 Maths chapter 10 straight line ?

No, you don't need to buy the NCERT solutions book for Class 11 Maths chapter 10 straight line. Here you will get NCERT Solutions for Class 11 Maths Chapter 10.

6. Do I need to buy NCERT exemplar solutions book for Class 11 Maths chapter 10 straight line ?

No, Here you will get NCERT Exemplar Solutions Class 11 Maths Chapter 10 for free.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

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Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11 - Straight Lines

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines Miscellaneous Exercise

Access Straight Lines Class 11 Chapter 10 Miscellaneous Exercise

More About NCERT Solutions for Class 11 Maths Chapter 10 Miscellaneous Exercise:-

Topic Covered in class 11 Maths ch 10 Miscellaneous Exercise Solutions

Also Read| Straight Lines Class 11 Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 10 Miscellaneous Exercise:-

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