Miscellaneous Exercise (Revised) - Chapter 10 - Straight Line - Ncert Solutions class 11 - Maths

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Chapter 9: Straight Line - NCERT Solutions for Class 11 Maths | Detailed Explanations

Miscellaneous Exercise Question 1.

Find the values of $k$ for which the line $(k-3) x-\left(4-k^2\right) y+k^2-7 k+6=0$ is:
(a) parallel to the $x$-axis
(b) parallel to the y-axis
(c) passing through the origin.

Answer.

Given: Equation of line $(k-3) x-\left(4-k^2\right) y+k^2-7 k+6=0$ can be written as
$
\begin{aligned}
& \left(4-\mathbf{k}^2\right) \mathbf{y}=(\mathbf{k}-3) \mathbf{x}+\mathbf{k}^2-7 \mathbf{k}+6 \\
& m=\frac{k-3}{4-k^2}
\end{aligned}
$
(a) If the line parallel to $x$-axis, then $m=0$
$
\begin{aligned}
& =>\frac{k-3}{4-k^2}=0 \\
& \Rightarrow \mathrm{k}-3=0 \\
& \Rightarrow \mathrm{k}=3
\end{aligned}
$
(b) If the line parallel to $y$-axis, then $\frac{1}{m}=0$
$
=>\frac{1}{\frac{k-3}{4-k^2}}=0
$

$\begin{aligned}
& \Rightarrow>\frac{4-k^2}{k-3}=0 \\
& \Rightarrow 4-\mathrm{k}^2=0 \\
& \Rightarrow \mathrm{k}^2=4
\end{aligned}$

$
\Rightarrow k= \pm 2
$
(c) If the line passes through origin then
$
\begin{aligned}
& \Rightarrow(k-3) \times 0-\left(4-k^2\right) \times 0+k^2-7 k+6=0 \\
& \Rightarrow k^2-7 k+6=0 \\
& \Rightarrow(k-1)(k-6)=0 \\
& \Rightarrow \mathrm{k}=1 \text { or } \mathrm{k}=6
\end{aligned}
$

Miscellaneous Exercise Question 2.

Find the equations of the lines which cut-off intercepts on the axes whose sum and 

product are 1 and -6 respectively.
Answer.

Let the intercepts be $\mathrm{a}$ and $\mathrm{b}$.
Then we have equation of line be $\frac{x}{a}+\frac{y}{b}=1$ and it is given that $a+b=1$ and $a b=-6$

Now $(a-b)^2=(a+b)^2-4 a b$
$
\begin{aligned}
& \Rightarrow(a-b)^2=(1)^2-4(-6) \\
& \Rightarrow(a-b)^2=1+24=25 \\
& \Rightarrow a-b= \pm 5
\end{aligned}
$

Solving $a+b=1$ and $a-b=5$,
we get $a=3, b=-2$
$\therefore$ Equation of the line is $\frac{x}{3}+\frac{y}{-2}=1$
$
\begin{aligned}
& \Rightarrow-2 x+3 y=-6 \\
& \Rightarrow 2 x-3 y=6
\end{aligned}
$

Solving $a+b=1$ and $a-b=-5$, we get $a=-2, b=3$
$\therefore$ Equation of the line is $\frac{x}{-2}+\frac{y}{3}=1$
$
\begin{aligned}
& \Rightarrow 3 x-2 y=-6 \\
& \Rightarrow-3 x+2 y=6
\end{aligned}
$

Hence required equations of the lines are $2 x-3 y=6$ and $-3 x+2 y=6$

Miscellaneous Exercise Question 3.

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

Answer.

Let point on $\mathrm{y}$ - axis be $(0, \mathrm{y})$
Given equation of line is $\frac{x}{3}+\frac{y}{4}=1$
$
\begin{aligned}
& \Rightarrow 4 x+3 y=12 \\
& \Rightarrow 4 x+3 y-12=0
\end{aligned}
$

We have perpendicular distance from $\left(x_1, y_1\right)$ to the line $a x+b y+c=0$ is
$
d=\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right|
$
$\therefore$ Perpendicular distance from point $(0, \mathrm{y})$ to $4 x+3 y-12=0=$
$
\left|\frac{4 \times 0+3 y-12}{\sqrt{(4)^2+(3)^2}}\right|=\left|\frac{3 y-12}{5}\right|
$

According to question $\left|\frac{3 y-12}{5}\right|=4$

$
\Rightarrow \frac{3 y-12}{5}= \pm 4
$

Taking $\frac{3 y-12}{5}=4$
$
\begin{aligned}
& \Rightarrow 3 y-12=20 \\
& \Rightarrow y=\frac{32}{3}
\end{aligned}
$

Taking $\frac{3 y-12}{5}=-4$
$
\begin{aligned}
& \Rightarrow 3 y-12=-20 \\
& \Rightarrow y=\frac{-8}{3}
\end{aligned}
$

Therefore, required points are $\left(0, \frac{32}{3}\right)$ and $\left(0, \frac{-8}{3}\right)$.
Miscellaneous Exercise Question 4.

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

Answer.

Equation of the line joining points $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$ is
$y-\sin \theta=\frac{\sin \phi-\sin \theta}{\cos \phi-\cos \theta}(x-\cos \theta)$, using Two-point form
$
\begin{aligned}
& \Rightarrow \\
& (\cos \phi-\cos \theta) y-\sin \theta \cos \phi+\sin \theta \cos \theta=(\sin \phi-\sin \theta) x-\sin \phi \cos \theta+\sin \theta \cos \theta \\
& \Rightarrow(\sin \phi-\sin \theta) x-(\cos \phi-\cos \theta) y+\sin \theta \cos \phi-\sin \phi \cos \theta=0
\end{aligned}
$

$
\Rightarrow(\sin \phi-\sin \theta) x-(\cos \phi-\cos \theta) y+\sin (\theta-\phi)=0
$

Now, perpendicular distance from $(0,0)$ to this line,
$
\begin{aligned}
& =\left|\frac{(\sin \phi-\sin \theta) \times 0-(\cos \phi-\cos \theta) \times 0+\sin (\theta-\phi)}{\sqrt{(\sin \phi-\sin \theta)^2+(\cos \phi-\cos \theta)^2}}\right| \text { since } d=\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right| \\
& =\left|\frac{\sin (\theta-\phi)}{\sqrt{\sin ^2 \phi+\sin ^2 \theta-2 \sin \phi \sin \theta+\cos ^2 \phi+\cos ^2 \theta-2 \cos \phi \cos \theta}}\right|
\end{aligned}
$

$\begin{aligned}
& =\left|\frac{\sin (\theta-\phi)}{\sqrt{1+1-2(\cos \theta \cos \phi+\sin \theta \sin \phi)}}\right| \text { since } \cos ^2 \theta+\sin ^2 \theta=1 \\
& =\left|\frac{\sin (\theta-\phi)}{\sqrt{2-2 \cos (\theta-\phi)}}\right| \text { using } \cos (\theta-\phi)=\cos \theta \cos \phi+\sin \theta \sin \phi \\
& =\left|\frac{\sin (\theta-\phi)}{\sqrt{2[1-\cos (\theta-\phi)]}}\right| \\
& =\left|\frac{\sin (\theta-\phi)}{\sqrt{2\left[2 \sin ^2\left(\frac{\theta-\phi}{2}\right)\right]}}\right| \text { since } 1-\cos \theta=2 \sin ^2 \frac{\theta}{2} \\
& =\left|\frac{\sin (\theta-\phi)}{2 \sin \left(\frac{\theta-\phi}{2}\right)}\right|
\end{aligned}$

Miscellaneous Exercise Question 5.

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

Answer.

The equation of any line parallel to $y$-axis is of the form $x=k, a$ constant
GIven lines are $x-7 y+5=0$
$
3 x+y=0
$

From (ii) we get $y=-3 x$
substituting y in equation(i), $x-7(-3 x)+5=0 \Rightarrow 22 x+5=0 \Rightarrow x=\frac{-5}{22}$ and $y=-3 x=-3 \times \frac{-5}{22}=\frac{15}{22}$
$\therefore$ Point of intersection of lines (i) and (ii) is $\left(\frac{-5}{22}, \frac{15}{22}\right)$

Since $\mathrm{x}=\mathrm{k}$,passes through the point $\left(\frac{-5}{22}, \frac{15}{22}\right)$ we get $k=\frac{-5}{22}$
Therefore the equation of required line is $x=\frac{-5}{22}$.
Miscellaneous Exercise Question 6.

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

Answer.

Given: Equation of line $\frac{x}{4}+\frac{y}{6}=1$
$
\Rightarrow 6 x+4 y-24=0
$

Slope of given line $=m=\frac{-6}{4}=\frac{-3}{2}$
Therefore slope of line perpendicular to given line $=\frac{-1}{m}=\frac{-1}{\frac{-3}{2}}=\frac{2}{3}$
Let the given line meet the $\mathrm{y}$-axis at $(0, \mathrm{y})$
Now by substituting $x=0$ in equation (i) , $4 y-24=0 \Rightarrow y=\frac{24}{4}=6$
$\therefore$ The given line meets the $y$ - axis at $(0,6)$.
Now the equation of a line with slope $\frac{2}{3}$ and passes through $(0,6)$ is

$
\begin{aligned}
& y-6=\frac{2}{3}(x-0) \\
& \Rightarrow 3 y-18=2 x \\
& \Rightarrow 2 x-3 y+18=0
\end{aligned}
$
$\therefore$ Equation of required line is $2 x-3 y+18=0$
Miscellaneous Exercise Question 7.

Find the area of the triangle formed by the lines $y-x=0, x+y=0$ and $x-k=0$.

Answer.

Given: Equations of lines are $y-x=0 \ldots .$. (i) $x+y=0$.....(ii)

And $x-k=0$ $\qquad$ ..(iii)

On solving eq. (i) and (ii), we get the point of intersection $\mathrm{C}=(0,0)$
On solving eq. (ii) and (iii), we get the point of intersection $\mathrm{A}=(k,-k)$
On solving eq. (i) and (iii), we get the point of intersection $\mathrm{B}=(k, k)$
We have the area of the triangle whose vertices are $\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is
$
\begin{aligned}
& \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| \\
& \therefore \text { Area } \quad \Delta \mathrm{ABC} \\
& =\frac{1}{2}|k(k-0)+k(0+k)+0(-k-k)|=\frac{1}{2}\left|k^2+k^2+0\right|=\frac{1}{2}\left|2 k^2\right|=k^2
\end{aligned}
$

Hence area of triangle formed by the three given lines is $\mathrm{k}^2$ sq. units
Miscellaneous Exercise Question 8.

Find the value of $\mathbf{p}$ so that three lines $3 x+y-2=0, p x+2 y-3=0$ and $2 x-y-3=0$ may intersect at one point.

Answer.

We know three lines $a_1 x+b_1 y+c_1 z=0, a_2 x+b_2 y+c_2 z=0 a n d a_3 x+b_3 y+c_3 z=0$ intersect at one point if $a_3\left(b_1 c_2-b_2 c_1\right)+b_3\left(c_1 a_2-c_2 a_1\right)+c_3\left(a_1 b_2-a_2 b_1\right)=0$

Given lines are $3 x+y-2=0, p x+2 y-3=0$ and $2 x-y-3=0$

$
\begin{aligned}
& \therefore 2[1 \times(-3)-2 \times(-2)]+(-1)[-2 \times p-(-3) \times 3]+(-3)[3 \times 2-p \times 1]=0 \\
& \Rightarrow 2[-3+4]-1[-2 p+9]-3[6-p]=0 \\
& \Rightarrow 2+2 p-9-18+3 p=0 \\
& \Rightarrow 5 p-25=0 \\
& \Rightarrow p=5
\end{aligned}
$

Hence the required value of $p$ is 5 .
Miscellaneous Exercise Question 9.

If three lines whose equations are $y=m_1 x+c_1 y=m_2 x+c_2$ and $y=m_3 x+c_3$ are concurrent., then show that $m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0$.

Answer.

We know three lines $a_1 x+b_1 y+c_1 z=0, a_2 x+b_2 y+c_2 z=0 a n d a_3 x+b_3 y+c_3 z=0$ are concurrent if $a_3\left(b_1 c_2-b_2 c_1\right)+b_3\left(c_1 a_2-c_2 a_1\right)+c_3\left(a_1 b_2-a_2 b_1\right)=0$

Given lines are $y=m_1 x+c_1, y=m_2 x+c_2$ and $y=m_3 x+c_3$
$\Rightarrow m_1 x-y+c_1=0, m_2 x-y+c_2=0$ and $m_3 x-y+c_3=0$
Since the lines are concurrent, we have
$
m_3\left(-1 . c_2+1 . c_1\right)-1\left(c_1 \cdot m_2-c_2 m_1\right)+c_3\left(m_1 \cdot-1+m_2 \cdot 1\right)=0
$

$
\begin{aligned}
& \Rightarrow m_3\left(-c_2+c_1\right)-c_1 \cdot m_2+c_2 m_1-c_3 m_1+c_3 m_2=0 \\
& \Rightarrow m_3\left(c_1-c_2\right)+m_2\left(c_3-c_1\right)+m_1\left(c_2-c_3\right)=0 \\
& \Rightarrow m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0
\end{aligned}
$
Miscellaneous Exercise Question 10.

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

Answer.

Let $m$ be the slope of required line which passes through point $(3,2)$,

Then the equation of required line is $y-2=m(x-3)$
The equation of given line $x-2 y=3$
$\Rightarrow y=\frac{x}{2}-\frac{3}{2}$ $\qquad$
$\therefore$ Slope of given line $=\frac{1}{2}$
We know that if $\theta$ is the acute angle between two lines with slopes $m_1$ and $m_2$ respectively then $\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|$
According to question, $\tan 45^{\circ}=\left|\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\right|$
$
\Rightarrow 1=\left|\frac{2 m-1}{2+m}\right|
$

$
\Rightarrow \frac{2 m-1}{2+m}= \pm 1
$

Taking $\frac{2 m-1}{2+m}=1$
$
\begin{aligned}
& \Rightarrow 2 m-1=2+m \\
& \Rightarrow m=3
\end{aligned}
$

Then the equation of required line is $y-2=3(x-3)$
$
\begin{aligned}
& \Rightarrow y-2=3 x-9 \\
& \Rightarrow 3 x-y-7=0
\end{aligned}
$

Taking $\frac{2 m-1}{2+m}=-1$
$
\begin{aligned}
& \Rightarrow 2 m-1=-2-m \\
& \Rightarrow m=\frac{-1}{3}
\end{aligned}
$

Then the equation of required line is $y-2=\frac{-1}{3}(x-3)$
$
\begin{aligned}
& \Rightarrow 3 y-6=-x+3 \\
& \Rightarrow x+3 y-9=0
\end{aligned}
$

Thus the equations of the lines are $3 x-y-7=0$ and $x+3 y-9=0$

Miscellaneous Exercise Question 11.

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

Answer.

Let the equal intercepts be a. Equation of a line in intercept form is $\frac{x}{a}+\frac{y}{a}=1 \Rightarrow x+y=a$

Given: The required equation passes through the point of intersection of the lines $4 x+7 y-3=0$ and $2 x-3 y+1=0$

Consider $4 x+7 y=3$ $\qquad$ (i) and $2 x-3 y=-1$ $\qquad$
Multiplying equation(ii) by 2 , we get $4 x-6 y=-2$. $\qquad$ (iii)
subtracting equation (iii) from equation(i), $13 y=5 \Rightarrow y=\frac{5}{13}$
substituting y in equation(ii) we get $2 x=-1+3 \cdot \frac{5}{13}=\frac{-13+15}{13}=\frac{2}{13} \Rightarrow x=\frac{1}{13}$
Hence the point of intersection of (i) and (ii) is $\left(\frac{1}{13}, \frac{5}{13}\right)$
Now since the required line $x+y=a$ passes through the point $\left(\frac{1}{13}, \frac{5}{13}\right)$ we have, $a=\frac{1}{13}+\frac{5}{13}=\frac{6}{13}$

$\therefore$ From equation(A) we get the required equation of the line is $x+y=\frac{6}{13} \Rightarrow 13 x+13 y=6$
Miscellaneous Exercise Question 12

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

Answer.

Let $m_1$ be the slope of required line which passes through $(0,0)$.
$\therefore$ Equation of line is $y-0=m_1(x-0)$
$
\Rightarrow y=m_1 x
$

Now, let $\theta$ be the angle between $y=m x+c$ and $y=m_1 x$
We know that if $\theta$ is the acute angle between two lines with slopes $m_1$ and $m_2$ respectively then $\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|$
$
\begin{aligned}
& \therefore \tan \theta=\left|\frac{m_1-m}{1+m_1 m}\right| \\
& \Rightarrow \tan \theta= \pm \frac{m_1-m}{1+m_1 m}
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \tan \theta=\frac{m_1-m}{1+m_1 m} \text { or } \tan \theta=-\frac{m_1-m}{1+m_1 m} \\
& \Rightarrow \tan \theta+m_1 m \tan \theta=m_1-m \text { or } \tan \theta+m_1 m \tan \theta=m-m_1 \\
& \Rightarrow m_1(1-m \tan \theta)=m+\tan \theta \text { or } m_1(1+m \tan \theta)=m-\tan \theta \\
& \Rightarrow m_1=\frac{m+\tan \theta}{1-m \tan \theta} \text { or } m_1=\frac{m-\tan \theta}{1+m \tan \theta}
\end{aligned}
$

Putting $m_1=\frac{m+\tan \theta}{1-m \tan \theta}$ in eq. (i), we get $y=\frac{m+\tan \theta}{1-m \tan \theta} x$.

Putting $m_1=\frac{m-\tan \theta}{1+m \tan \theta}$ in eq. (i), we get $y=\frac{m-\tan \theta}{1+m \tan \theta} x$.
Therefore from (ii) and (iii) we get $\frac{y}{x}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta}$
Miscellaneous Exercise Question 13.

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

Answer.

Given: Equation of line $x+y-4=0$
Let the given line divide the line joining $\mathrm{A}(-1,1)$ and $\mathrm{B}(5,7)$ in the ratio $k: 1$ at a point $\mathrm{C}$.

$\therefore$ Using section formula we have coordinates of $\mathrm{C}$ are $\left(\frac{k(5)+1(-1)}{k+1}, \frac{k(7)+1(1)}{k+1}\right)$
$
=\left(\frac{5 k-1}{k+1}, \frac{7 k+1}{k+1}\right)
$

Since the point $\mathrm{C}$ lies on the given line.
$
\begin{aligned}
& \therefore \frac{5 k-1}{k+1}+\frac{7 k+1}{k+1}-4=0 \\
& \Rightarrow 5 k-1+7 k+1-4 k-4=0 \\
& \Rightarrow 8 k=4 \\
& \Rightarrow k=\frac{1}{2}
\end{aligned}
$

Therefore, the required ratio is $1: 2$.

Miscellaneous Exercise Question 14.

Find the distance of the line $4 x+7 y+5=0$ from the point $(1,2)$ along the line $2 x-1$

Answer.

First we have to find the point of intersection of given lines $4 x+7 y+5=0$ and $2 x-y=0$ (ii)

From(ii) we get $y=2 x$
Substituting $y=2 x$ in equation(i), we get $18 x+5=0 \Rightarrow x=\frac{-5}{18}$
$
\therefore y=2 x=2\left(\frac{-5}{18}\right)=\frac{-5}{9}
$

Hence point of intersection of (i) and (ii) is $\mathrm{A}\left(\frac{-5}{18}, \frac{-5}{9}\right)$
$\therefore$ Distance between the points A $\left(\frac{-5}{18}=\frac{-5}{9}\right)$ and $\mathrm{B}(1,2)$
$
\begin{aligned}
& =\sqrt{\left(1+\frac{5}{18}\right)^2+\left(2+\frac{5}{9}\right)^2} \\
& =\sqrt[1]{\left(\frac{23}{18}\right)^2+\left(\frac{23}{9}\right)^2} \\
& =\frac{23}{18} \sqrt[4]{(1)^2+(2)^2}=\frac{23}{18} \sqrt{5} \text { units }
\end{aligned}
$
Miscellaneous Exercise Question 15.

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

Answer.

Let the required equation of the line be $y-y_0=m\left(x-x_0\right)$
Since equation (i) passes through the point $(-1,2)$, we get $y-2=m(x+1)$
$
\Rightarrow m x-y+(m+2)=0
$

Equation of the given line is $x+y=4$ $\qquad$
From(iii) we get $y=4-x$,substituting this in(ii),we get
$
\begin{aligned}
& m x-(4-x)+m+2=0 \Rightarrow x(m+1)+(m-2)=0 \Rightarrow x=\frac{2-m}{m+1} \\
& \therefore y=4-\left(\frac{2-m}{m+1}\right)=\frac{4 m+4-2+m}{m+1}=\frac{5 m+2}{m+1}
\end{aligned}
$

Hence we get point of intersection of (ii) and (iii) is $\left(\frac{2-m}{m+1}, \frac{5 m+2}{m+1}\right)$
Given distance between $(-1,2)$ and $\left(\frac{2-m}{m+1}, \frac{5 m+2}{m+1}\right)$ is 3
Now using distance formula we get $\sqrt{\left(\frac{2-m}{m+1}+1\right)^2+\left(\frac{5 m+2}{m+1}-2\right)^2}=3$
squaring on both sides $\left(\frac{2-m+m+1}{m+1}\right)^2+\left(\frac{5 m+2-2 m-2}{m+1}\right)^2=9$
$
\begin{aligned}
& \Rightarrow\left(\frac{3}{m+1}\right)^2+\left(\frac{3 m}{m+1}\right)^2=9 \Rightarrow 9\left[\frac{1+m^2}{(m+1)^2}\right]=9 \\
& \Rightarrow 1+m^2=(m+1)^2 \Rightarrow 1+m^2=m^2+1+2 m
\end{aligned}
$

$
\Rightarrow 2 m=0 \Rightarrow m=0
$

Since the slope is zero,we have the required line is parallel to $x$-axis .
Miscellaneous Exercise Question 16.

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

Answer.

Let ABC be a right angled triangle with diagonal AC.

Then we have the legs of the triangle (perpendicular sides) are BA and BC

Let the slope of $\mathrm{AB}$ is $\mathrm{m}$,then since $\mathrm{AB}$ and $\mathrm{BC}$ are perpendicular to each other we have, Slope of $\mathrm{BC}=\frac{-1}{m}$

We have the equation of a line passing through $\left(x_0, y_0\right)$ and slope $\mathrm{m}$ is $y-y_0=m\left(x-x_0\right)$ Now since $A B$ passes through $A(1,3)$ and have slope $m$ equation of $A B$ is
$
y-3=m(x-1)
$

Also $\mathrm{BC}$ is a line through $\mathrm{C}(-4,1)$. with slope $\frac{-1}{m}$ hence it'sequation is
$
y-1=\frac{-1}{m}(x+4)
$

By putting different values for $m$ in equations (i) and (ii) we get different equations for BA and BC

In particular for $\mathrm{m}=0$,equation of $\mathrm{AB}$ is $\mathrm{y}=3$ [ which is a line parallel to $\mathrm{X}$-axis] and equation of $\mathrm{BC}$ is $\mathrm{x}=-4$ [ which is a line parallel to Y-axis]
Miscellaneous Exercise Question 17.

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

Answer.

Let the image of the point $\mathrm{A}(3,8)$ in the line mirror $\mathrm{DE}$ be $\mathrm{C}(\alpha, \beta)$. Then $\mathrm{AC}$ is perpendicular bisector of DE.

$\therefore$ The coordinates of point B are $\left(\frac{\alpha+3}{2}=\frac{\beta+8}{2}\right)$
Since point B lies on the line $x+3 y=7$.
$
\begin{aligned}
& \therefore \frac{\alpha+3}{2}+\frac{3(\beta+8)}{2}=7 \\
& \Rightarrow \alpha+3+3 \beta+24=14 \\
& \Rightarrow \alpha+3 \beta+13=0 \ldots .(\mathrm{i})
\end{aligned}
$

Since AC is perpendicular on DE.
$\therefore$ Slope of AC $x$ Slope of $D E=-1$
$
\begin{aligned}
& \Rightarrow \frac{\beta-8}{\alpha-3} \times \frac{-1}{3}=-1 \\
& \Rightarrow \beta-8=3 \alpha-9 \\
& \Rightarrow 3 \alpha-\beta-1=0 \ldots
\end{aligned}
$

Solving eq. (i) and (ii), we get $\alpha=-1$ and $\beta=-4$
Therefore, the image of point $(3,8)$ is $(-1,-4)$.

Miscellaneous Exercise Question 18.

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

Answer.

Given lines are $y=3 x+1$ $\qquad$
$
2 y=x+3
$
$\qquad$ (ii) and $y=m x+4$ $\qquad$
Now we have slopes of lines (i),(ii) and (iii) are 3,1/2,m respectively
We know that if $\theta$ is the acute angle between two lines with slopes $m_1$ and $m_2$ respectively then $\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|$

Let $\theta$ be the angle which the line $y=m x+4$ makes with the line $y=3 x+1$ and $2 y=x+3$.
$\therefore \tan \theta=\left|\frac{m-3}{1+3 m}\right|$ and $\tan \theta=\left|\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\right|=\left|\frac{2 m-1}{2+m}\right|$
Given that lines $y=3 x+1$ and $2 y=x+3$ are equally inclined to the line $y=m x+4$
$
\begin{aligned}
& \Rightarrow\left|\frac{m-3}{1+3 m}\right|=\left|\frac{2 m-1}{2+m}\right| \\
& \Rightarrow \frac{m-3}{1+3 m}= \pm \frac{2 m-1}{2+m} \\
& \Rightarrow m^2-m-6= \pm\left(6 m^2-m-1\right)
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow 5 m^2+5=0 \text { or } 7 m^2-2 m-7=0 \\
& \Rightarrow m=\frac{1 \pm 5 \sqrt{2}}{7}
\end{aligned}
$
Miscellaneous Exercise Question 19.

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

Answer.

Given: Equations of lines are $x+y-5=0$

and $3 x-2 y+7=0$
Perpendicular distance of point $\mathrm{P}(x, y)$ from line (i)
$
=\left|\frac{x+y-5}{\sqrt{(1)^2+(1)^2}}\right|=\left|\frac{x+y-5}{\sqrt{2}}\right|
$

Perpendicular distance of point $\mathrm{P}(x, y)$ from line (ii)
$
\left|\frac{3 x-2 y+7}{\sqrt{(3)^2+(-2)^2}}\right|=\left|\frac{3 x-2 y+7}{\sqrt{13}}\right|
$

According to question, $\left|\frac{x+y-5}{\sqrt{2}}\right|+\left|\frac{3 x-2 y+7}{\sqrt{13}}\right|=10$
$
\Rightarrow|\sqrt{13}(x+y-5)|+|\sqrt{2}(3 x-2 y+7)|=10 \sqrt{26}
$

When $x+y-5 \geq 0$ and $3 x-2 y+7 \geq 0$
Then $|\sqrt{13}(x+y-5)|+|\sqrt{2}(3 x-2 y+7)|=10 \sqrt{26}$
$
\Rightarrow \sqrt{13} x+\sqrt{13} y-5 \sqrt{13}+3 \sqrt{2} x-2 \sqrt{2} y+7 \sqrt{2}-10 \sqrt{26}=0
$
$\Rightarrow(\sqrt{13}+3 \sqrt{2}) x+(\sqrt{13}-2 \sqrt{2}) y-5 \sqrt{13}+7 \sqrt{2}-10 \sqrt{26}=0$, which represent a line.

Similarly,we can get the equation of a line for any signs ( positive or negative ) of $x+y-5$ and $3 x-2 y+7$

Therefore, we can say that $\mathrm{P}(x, y)$ must move on a line.
Miscellaneous Exercise Question 20.

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

Answer.

The equations of parallel lines are $9 x+6-7=0$ and $3 x+2 y+6=0$

Let $\mathrm{A}\left(x_1, y_1\right)$ be any point which is equidistant from the parallel lines.
$
\begin{aligned}
& \therefore\left|\frac{9 x_1+6 y_1-7}{\sqrt{(9)^2+(6)^2}}\right|=\left|\frac{3 x_1+2 y_1+6}{\sqrt{(3)^2+(2)^2}}\right| \\
& \Rightarrow \frac{9 x_1+6 y_1-7}{3 \sqrt{13}}= \pm \frac{3 x_1+2 y_1+6}{\sqrt{13}}
\end{aligned}
$

Consider the case $\frac{9 x_1+6 y_1-7}{3 \sqrt{13}}=+\frac{3 x_1+2 y_1+6}{\sqrt{13}}$
$\Rightarrow 9 x_1+6 y_1-7=3\left(3 x_1+2 y_1+6\right) \Rightarrow-7=18$ which leads to some wrong conclusion
so this case is not possible
Now consider $\frac{9 x_1+6 y_1-7}{3 \sqrt{13}}=-\frac{3 x_1+2 y_1+6}{\sqrt{13}}$
$
\begin{aligned}
& \Rightarrow 9 x_1+6 y_1-7=-9 x_1-6 y_1-18 \\
& \Rightarrow 18 x_1+12 y_1+11=0
\end{aligned}
$

Therefore, the required line is $18 x+12 y+11=0$.

Miscellaneous Exercise Question 21.

A ray of light passing through the point $A$ and the reflected ray passes 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.

Let the coordinates of the point $\mathrm{A}$ be $(\mathrm{x}, 0$ )
Also let BA be the incident ray and $\mathrm{AC}$ be the reflected ray and $\mathrm{AD}$ be the normal

We have angle of incidence $=$ angle of reflection $\Rightarrow \angle B A D=\angle C A D$
Let $\angle B A D=\angle C A D=\theta$
Also let $\angle C A X=\phi$
Then we have $\phi+\theta=90^{\circ} \Rightarrow \theta=90^{\circ}-\phi$
Also $\angle O A B=180^{\circ}-(\phi+2 \theta)=180^{\circ}-\left[\phi+2\left(90^{\circ}-\phi\right)\right]=180^{\circ}-\left(180^{\circ}-\phi\right)=\phi$
Now, for line AC $\tan \phi=\frac{3-0}{5-x}$
$\Rightarrow \tan \phi=\frac{3}{5-x}$
For line BA $\tan \left(180^{\circ}-\phi\right)=\frac{2-0}{1-x}$
$
\begin{aligned}
& \Rightarrow-\tan \phi=\frac{2}{1-x} \\
& \Rightarrow \tan \phi=\frac{-2}{1-x} \ldots . .
\end{aligned}
$

From eq. (i) and (ii), $\frac{3}{5-x}=\frac{-2}{1-x}$
$
\Rightarrow 3-3 x=-10+2 x
$

$\Rightarrow-5 x=-13$

$
\Rightarrow x=\frac{13}{5}
$

Therefore, coordinates of point A are $\left(\frac{13}{5}, 0\right)$.

Miscellaneous Exercise Question 22.

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

Ans. Let $p_1$ and $p_2$ be the length of perpendiculars from $\left(\sqrt{a^2-b^2}=0\right)$ and $\left(-\sqrt{a^2-b^2}=0\right)$ to the line $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$

$=\left|\frac{\frac{\sqrt{a^2-b^2} \cos \theta}{a}-1}{\sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}}\right|$

$\begin{aligned}
& =a^2-\left(a^2-b^2\right) \cos ^2 \theta \times \frac{b^2}{a^2-\left(a^2-b^2\right) \cos ^2 \theta}=b^2 \\
&
\end{aligned}$

Miscellaneous Exercise Question 24.

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

Answer.

Consider $2 x-3 y+4=0$ $\qquad$ (i) and $3 x+4 y-5=0$ $\qquad$
The point of intersection of lines (i) and (ii) is given by $A\left(\frac{22}{17}, \frac{-2}{17}\right)$
It is given that the person is standing at $\mathrm{A}$ and wants to reach the path whose equation is $6 x-7 y+8=0$ $\qquad$ (iii) in the least time

We have the shortest path from point A to line(iii) is the line $\mathrm{AB}$ which is the perpendicular distance from $A\left(\frac{-1}{17}, \frac{22}{17}\right)$ to equation (iii)

$\therefore$ Slope of line $6 x-7 y+8=0$ is $\frac{6}{7}$
$\therefore$ Slope of required line $\mathrm{AB}$ is $\frac{-7}{6}$
Therefore, the equation of $\mathrm{AB}$ is $y-y_0=m\left(x-x_0\right)$
$
\Rightarrow y-\frac{22}{17}=\frac{-7}{6}\left(x+\frac{1}{17}\right)
$

$
\begin{aligned}
& \Rightarrow \frac{17 y-22}{17}=\frac{-7}{6}\left(\frac{17 x+1}{17}\right) \\
& \Rightarrow 6(17 y-22)=-7(17 x+1) \\
& \Rightarrow 102 y-132=-119 x-7 \\
& \Rightarrow 119 x+102 y=125
\end{aligned}
$

Hence the equation of the path that the person should follow is $119 x+102 y=125$