Exercise 9.3 (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

Ex 9.3 Question 1.

Reduce the following equations into slope-intercept form and find their slopes and the $y$ - intercepts.
(i) $x+7 y=0$
(ii) $6 x+3 y-5=0$
(iii) $y=0$

Answer.

(i) Given: $x+7 y=0$
$
\begin{aligned}
& \Rightarrow 7 y=-x \\
& \Rightarrow y=\frac{-1}{7} x+0
\end{aligned}
$

Comparing with $y=m x+c$, we have $m=\frac{-1}{7}$ and $c=0$
(ii) Given: $6 x+3 y-5=0$
$
\begin{aligned}
& \Rightarrow 3 y=-6 x+5 \\
& \Rightarrow y=-2 x+\frac{5}{3}
\end{aligned}
$

Comparing with $y=m x+c$. we have $m=-2$ and $c=\frac{5}{3}$

(iii) Given: $y=0$
$
\Rightarrow y=0 x+0
$

Comparing with $y=m x+c$, we have $m=0$ and $c=0$
Ex 9.3 Question 2.

Reduce the following equations into intercept form and find their intercepts on the axes:
(i) $3 x+2 y-12=0$
(ii) $4 x-3 y=6$
(iii) $3 y+2=0$

Answer.

(i) Given: $3 x+2 y-12=0$
$
\begin{aligned}
& \Rightarrow 3 x+2 y=12 \\
& \Rightarrow \frac{3 x}{12}+\frac{2 y}{12}=1 \\
& \Rightarrow \frac{x}{4}+\frac{y}{6}=1
\end{aligned}
$

Comparing with $\frac{x}{a}+\frac{y}{b}=1$, we have $a=4$ and $b=6$
(ii) Given: $4 x-3 y=6$

$
\begin{aligned}
& \Rightarrow \frac{4 x}{6}-\frac{3 y}{6}=1 \\
& \Rightarrow \frac{x}{3 / 2}+\frac{y}{-2}=1
\end{aligned}
$

Comparing with $\frac{x}{a}+\frac{y}{b}=1$, we have $a=\frac{3}{2}$ and $b=-2$
(iii) Given: $3 y+2=0$
$
\Rightarrow 3 y=-2
$

$
\begin{aligned}
& \Rightarrow \frac{3 y}{-2}=1 \\
& \Rightarrow \frac{y}{\frac{-2}{3}}=1
\end{aligned}
$

Comparing with $\frac{x}{a}+\frac{y}{b}=1$, we have $b=\frac{-2}{3}$ and no inercept with $\mathrm{x}$-axis.

Ex 9.3 Question 3.

Find the distance of the point $(-1,1)$ from the line $12(x+6)=5(y-2)$.

Answer.

Given equation of the line is $12(x+6)=5(y-2)$
$
\begin{aligned}
& \Rightarrow 12 x+72=5 y-10 \\
& \Rightarrow 12 x-5 y+82=0
\end{aligned}
$

We have the distance of a given point $\left(x_1, y_1\right)$ from a given line $\mathrm{Ax}+\mathrm{BY}+\mathrm{C}=0$ is
$
d=\left|\frac{A x_1+B y_1+C}{\sqrt{A^2+B^2}}\right|
$
$\therefore$ Perpendicular distance of the point $(-1,1)$ from the line $12 x-5 y+82=0$ is
$
\left|\frac{12(-1)-5(1)+82}{\sqrt{(12)^2+(-5)^2}}\right|=\left|\frac{-12-5+82}{\sqrt{144+25}}\right|=\left|\frac{65}{13}\right|=5 \text { units }
$
Ex 9.3 Question 4.

Find the points on the $x$-axis, whose distances from the line $\frac{x}{3}+\frac{y}{4}=1$ are 4 units.

Answer.

Let the coordinates of the point on $x$-axis be $(\alpha, 0)$.

Given line is $\frac{x}{3}+\frac{y}{4}=1 \Rightarrow 4 x+3 y-12=0$
We have the distance of a given point $\left(x_1, y_1\right)$ from a given line $\mathrm{Ax}+\mathrm{BY}+\mathrm{C}=0$ is
$
d=\left|\frac{A x_1+B y_1+C}{\sqrt{A^2+B^2}}\right|
$

Therefore perpendicular distance of the point $(\alpha, 0)$ from the line $4 x+3 y-12=0$ is
$
\left|\frac{4(\alpha)+3(0)-12}{\sqrt{(4)^2+(3)^2}}\right|=\left|\frac{4 \alpha-12}{\sqrt{16+9}}\right|=\left|\frac{4 \alpha-12}{5}\right|
$

According to question, $\left|\frac{4 \alpha-12}{5}\right|=4$
$
\begin{aligned}
& \Rightarrow \frac{4 \alpha-12}{5}= \pm 4 \\
& \Rightarrow \frac{4 \alpha-12}{5}=4 \text { or } \frac{4 \alpha-12}{5}=-4 \\
& \Rightarrow 4 \alpha-12=20 \text { or } 4 \alpha-12=-20 \\
& \Rightarrow 4 \alpha=32 \text { or } 4 \alpha=-8
\end{aligned}
$

$
\Rightarrow \alpha=8 \text { or } \alpha=-2
$

Therefore, the points on $x$-axis are $(8,0)$ and $(-2,0)$.
Ex 9.3 Question 5.

Find the distance between parallel lines:
(i) $15 x+8 y-34=0$ and $15 x+8 y+31=0$
(ii) $l(x+y)+p=0$ and $l(x+y)-r=0$

Answer.

(i) Given: Two equations $15 x+8 y-34=0$ and $15 x+8 y+31=0$
Here, $a=15, b=8, c_1=-34$ and $c_2=31$
Distance between two parallel lines $(d)=\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=\frac{|-34-31|}{\sqrt{(15)^2+(8)^2}}$
$
=\frac{|-65|}{\sqrt{225+64}}=\frac{65}{\sqrt{289}}=\frac{65}{17} \text { units }
$
(ii) Given: Two equations $b x+b y+p=0$ and $b x+b y-r=0$

Here, $a=l, b=l, c_1=p$ and $c_2=-r$
Distance between two parallel lines $(d)=\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=\frac{|p+r|}{\sqrt{(l)^2+(l)^2}}$
$
=\frac{|p+r|}{\sqrt{2 l^2}}=\frac{1}{\sqrt{2}}\left|\frac{p+r}{l}\right| \text { units }
$

Ex 9.3 Question 6.

Find equation of the line parallel to the line $3 x-4 y+2=0$ and passing through the point $(-2,3)$

Answr.

 We have equation of any line parallel to $A x+B Y+C=0$ is of the form $A x+B y+K=0, K$ is a constant
$\therefore$ Equation of a line which is parallel to the line $3 x-4 y+2=0$ is $3 x-4 y+k=0$.
Since the line passes through the point $(-2,3)$.
we have $3 \times(-2)-4 \times 3+k=0$
$
\begin{aligned}
& \Rightarrow-6-12+k=0 \\
& \Rightarrow k=18
\end{aligned}
$

Therefore, the equation of required line is $3 x-4 y+18=0$.
Ex 9.3 Question 7.

Find the equation of the line perpendicular to the line $x-7 y+5=0$ and having $x$ intercept 3 .

Answer.

We have equation of any line perpendicular to $\mathrm{Ax}+\mathrm{BY}+\mathrm{C}=0$ is of the form $\mathrm{Bx}-\mathrm{Ay}+\mathrm{K}=0, \mathrm{~K}$ is a constant
$\therefore$ Equation of a line which is perpendicular to the line $x-7 y+5=0$ is $7 x+y+k=0$.
Since the line passes through the point $(3,0)$.
we have $7 \times 3+0+k=0$
$
\begin{aligned}
& \Rightarrow 21+k=0 \\
& \Rightarrow k=-21
\end{aligned}
$

Therefore, the equation of required line is $7 x+y-21=0$.
Ex 9.3 Question 8.

Find the angles between the lines $\sqrt{3} x+y=1$ and $x+\sqrt{3} y=1$.

Answer.

Given: $\sqrt{3} x+y=1$
$
\Rightarrow y=-\sqrt{3} x+1
$

$
\therefore m_1=-\sqrt{3}
$

Also $x+\sqrt{3} y=1$
$
\begin{aligned}
& \Rightarrow y=-\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}} \\
& \therefore m_2=-\frac{1}{\sqrt{3}}
\end{aligned}
$

An acute angle $\theta$ between the two lines is given by $\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$

$
\begin{aligned}
& \therefore \tan \theta=\left|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)}\right| \\
& =\left|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}\right|=\left|\frac{-2}{\sqrt{3}} \times \frac{1}{2}\right|=\frac{1}{\sqrt{3}} \\
& \Rightarrow \tan \theta=\tan 30^{\circ} \text { and } \tan \left(180^{\circ}-30^{\circ}\right)=\tan 150^{\circ} \\
& \Rightarrow \theta=30^{\circ} \text { and } 150^{\circ}
\end{aligned}
$
Ex 9.3 Question 9.

The line through the points $(h, 3)$ and $(4,1)$ intersects the line $7 x-9 y-19=0$ at right angle. Find the value of $h$.

Aswer.

Slope of the line passing through the points $(h, 3)$ and $(4,1)=\frac{1-3}{4-h}=\frac{-2}{4-h}$
Also slope of the line $7 x-9 y-19=0$ is $\frac{7}{9}$
We have if two lines are perpendicular to each other then products of their slopes $=-1$

$\begin{aligned}
& \therefore \frac{-2}{4-h} \times \frac{7}{9}=-1 \\
& \Rightarrow \frac{-14}{36-9 h}=-1 \\
& \Rightarrow-14=-36+9 h \\
& \Rightarrow 9 h=36-14 \\
& \Rightarrow h=\frac{22}{9}
\end{aligned}$

Ex 9.3 Question 10.

Prove that the line through the point $\left(x_1, y_1\right)$ and parallel to the line $\mathrm{A} x+\mathrm{B} y+\mathrm{C}=0$ is $\mathrm{A}\left(x-x_1\right)+\mathrm{B}\left(y-y_1\right)=0$.

Answer.

Equation of any line parallel to the line $A x+B y+C=0$ is $A x+B y+K=0$
Since line (i) passes through $\left(x_1, y_1\right)$, we get $\mathrm{A} x_1+\mathrm{B} y_1+\mathrm{K}=0$
Subtracting eq. (ii) from eq. (i), we have $A\left(x-x_1\right)+B\left(y-y_1\right)=0$
Ex 9.3 Question 11.

Two lines passing through the point $(2,3)$ intersects each other at an angle of $60^{\circ}$. If slope of one line is 2 , find equation of the other line.

Answer.

Given: $m_1=2$ and $\theta=60^{\circ}$
An acute angle $\theta$ between the two lines is given by $\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$
$
\begin{aligned}
& \Rightarrow \tan 60^{\circ}=\left|\frac{2-m_2}{1+2 m_2}\right| \\
& \Rightarrow \sqrt{3}=\left|\frac{2-m_2}{1+2 m_2}\right|
\end{aligned}
$

$
\Rightarrow \frac{2-m_2}{1+2 m_2}= \pm \sqrt{3}
$

Taking $\frac{2-m_2}{1+2 m_2}=\sqrt{3}$
$
\begin{aligned}
& \Rightarrow 2-m_2=\sqrt{3}+2 \sqrt{3} m_2 \\
& \Rightarrow(2 \sqrt{3}+1) m_2=2-\sqrt{3} \\
& \Rightarrow m_2=\frac{2-\sqrt{3}}{2 \sqrt{3}+1}
\end{aligned}
$

$\therefore$ Equation of required line is $y-3=\frac{2-\sqrt{3}}{2 \sqrt{3}+1}(x-2)$
$
\begin{aligned}
& \Rightarrow(2 \sqrt{3}+1) y-6 \sqrt{3}-3=(2-\sqrt{3}) x-4+2 \sqrt{3} \\
& \Rightarrow(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-4+2 \sqrt{3}+6 \sqrt{3}+3 \\
& \Rightarrow(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=8 \sqrt{3}-1
\end{aligned}
$

Taking $\frac{2-m_2}{1+2 m_2}=-\sqrt{3}$
$
\begin{aligned}
& \Rightarrow 2-m_2=-\sqrt{3}-2 \sqrt{3} m_2 \\
& \Rightarrow(2 \sqrt{3}-1) m_2=-(2+\sqrt{3}) \\
& \Rightarrow m_2=\frac{-(2+\sqrt{3})}{2 \sqrt{3}-1}
\end{aligned}
$
$\therefore$ Equation of required line is $y-3=\frac{-(2+\sqrt{3})}{2 \sqrt{3}-1}(x-2)$

$
\begin{aligned}
& \Rightarrow(2 \sqrt{3}-1) y-6 \sqrt{3}+3=-(2+\sqrt{3}) x+4+2 \sqrt{3} \\
& \Rightarrow(2+\sqrt{3}) x+(2 \sqrt{3}-1) y=4+2 \sqrt{3}+6 \sqrt{3}-3 \\
& \Rightarrow(2+\sqrt{3}) x+(2 \sqrt{3}-1) y=8 \sqrt{3}+1
\end{aligned}
$
Ex 9.3 Question 13.

Find the equation of the right bisector of the line segment joining the points (3, 4) and $(-1,2)$.

Answer.

Let $\mathrm{A}(3,4)$ and $\mathrm{B}(-1,2)$ be the given points
Midpoint of the line segment joining the points $A$ and $B=\left(\frac{3-1}{2}, \frac{4+2}{2}\right)=(1,3)$
Slope of the line joining points $\mathrm{A}(3,4)$ and $B(-1,2)=\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}$
Since the required line is perpendiculat to the line $A B$,slope of the required line is -2
We have the equation of a line passing through $\left(x_{\circ}, y_{\circ}\right)$ and slope $\mathrm{m}$ is $y-y_0=m\left(x-x_0\right)$
Since the required line passes through point $(1,3)$ and having slope -2 .
We have , equation of the required line is $y-3=-2(x-1)$
$
\begin{aligned}
& \Rightarrow y-3=-2 x+2 \\
& \Rightarrow 2 x+y-5=0
\end{aligned}
$

We have , equation of the required line is $y-3=-2(x-1)$
$
\begin{aligned}
& \Rightarrow y-3=-2 x+2 \\
& \Rightarrow 2 x+y-5=0
\end{aligned}
$
Ex 9.3 Question 14.

Find the coordinates of the foot of perpendicular from the point $(-1,3)$ to the line $3 x$

Answer.

Let $\mathrm{Q}$ be the foot of perpendicular drawn from $\mathrm{P}(-1,3)$ on the line $3 x-4 y-16=0$

$\therefore$ Equation of a line perpendicular to $3 x-4 y-16=0$ is $4 x+3 y+k=0$
Since the line passes through $(-1,3)$
$
\therefore 4 \times(-1)+3 \times 3+k=0
$

$
\begin{aligned}
& \Rightarrow-4+9+k=0 \\
& \Rightarrow k=-5
\end{aligned}
$

Therefore, $Q$ is a point of intersection of the lines $3 x-4 y-16=0$ and $4 x+3 y-5=0$
Solving both the equations, we have $x=\frac{68}{25}$ and $y=\frac{-49}{25}$
Therefore, coordinates of foot of perpendicular are $\left(\frac{68}{25}, \frac{-49}{25}\right)$.
Ex 9.3 Question 15.

The perpendicular from the origin to the line $y=m x+c$ meets it at the point $(-1,2)$. Find the value of $m$ and $c$.

Answer.

Given equation of line is $y=m x+c$.
Since the perpendicular from the origin meets the above line (i) at $(-1,2)$, we have the line joining $(0,0)$ and $(-1,2)$ is perpendicular to the line (i)
Now slope of line joining $(0,0)$ and $(-1,2)=\frac{2-0}{-1-0}=-2$
and slope of given line $=\mathrm{m}$
But we have if two lines are perpendicular the product of their slopes is equal to -1 .
$
\therefore(-2) m=-1 \Rightarrow m=\frac{-1}{-2}=\frac{1}{2}
$

Since the point $(-1,2)$ lies on the line(i) we get $2=m(-1)+c \Rightarrow 2=\frac{1}{2}(-1)+c$
$
\Rightarrow c=2+\frac{1}{2}=\frac{5}{2}
$

Hence we have $m=\frac{1}{2}$ and $c=\frac{5}{2}$
Ex 9.3 Question 16.

If $p$ and $q$ are the lengths of perpendiculars from the origin to the line $x \cos \theta-y \sin \theta=k \cos 2 \theta$ and $x \sec \theta+y \operatorname{cosec} \theta=k$ respectively, prove that $p^2+4$

Answer.

We have the length of the perpendicular from of a given point $\left(x_1, y_1\right)$ to a given line $\mathrm{Ax}+\mathrm{BY}+\mathrm{C}=0$ is $d=\left|\frac{A x_1+B y_1+C}{\sqrt{A^2+B^2}}\right|$

Length of perpendicular from origin to line $x \cos \theta-y \sin \theta-k \cos 2 \theta=0$ is
$
\begin{aligned}
& p=\left|\frac{0 \times \cos \theta-0 \times \sin \theta-k \cos 2 \theta}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}}\right| \\
& =\left|\frac{-k \cos 2 \theta}{1}\right|=k \cos 2 \theta
\end{aligned}
$

And Length of perpendicular from origin to line $x \sec \theta+y \operatorname{cosec} \theta-k=0$ is
$
\begin{aligned}
& \text { since } \cos ^2 \theta+\sin ^2 \theta=1 \\
& =|-k \sin \theta \cos \theta|=\frac{k}{2} \sin 2 \theta \\
& {[\sin 2 \theta=2 \sin \theta \cos \theta]} \\
&
\end{aligned}
$

Now, $p^2+4 q^2=(k \cos 2 \theta)^2+4\left(\frac{k}{2} \sin 2 \theta\right)^2$
$
\begin{aligned}
& =k^2 \cos ^2 2 \theta+\frac{4}{4} k^2 \sin ^2 2 \theta \quad \text { since } \cos ^2 \theta+\sin ^2 \theta=1 \\
& =k^2\left(\cos ^2 2 \theta+\sin ^2 2 \theta\right)=k^2
\end{aligned}
$

Therefore $p^2+4 q^2=k^2$.

Ex 9.3 Question 17.

In the triangle $\mathrm{ABC}$ with vertices $\mathrm{A}(2,3), \mathrm{B}(4,-1)$ and $\mathrm{C}(1,2)$, find the equation and length of altitude from the vertex $A$.

Answer.

Slope of $\mathrm{BC}=\frac{2-(-1)}{1-4}=\frac{2+1}{-3}=\frac{3}{-3}=-1$

Since $A D \perp B C$, therefore slope of $A D=1$
We have the equation of a line passing through $\left(x_{\circ}, y_0\right)$ and slope $\mathrm{m}$ is $y-y_0=m\left(x-x_0\right)$
$\therefore$ Equation of altitude $\mathrm{AD}$ is
$
\begin{aligned}
& y-3=1(x-2) \\
& \Rightarrow x-y+1=0
\end{aligned}
$

And Equation of $\mathrm{BC}$ is
$
\begin{aligned}
& y+1=-1(x-4) \\
& \Rightarrow x+y-3=0
\end{aligned}
$

We have the length of the perpendicular from of a given point $\left(x_1, y_1\right)$ to a given line $\mathrm{Ax}+\mathrm{BY}+\mathrm{C}=0$ is $d=\left|\frac{A x_1+B y_1+C}{\sqrt{A^2+B^2}}\right|$
$\therefore$ Length of $\mathrm{AD}=$ Perpendicular distance from $(2,3)$ to the line $\mathrm{BC}=\left|\frac{2+3-3}{\sqrt{(1)^2+(1)^2}}\right|$

$
=\left|\frac{2}{\sqrt{2}}\right|=\sqrt{2} \text { units }
$
$\therefore$ Equation of the altitude from A is $x-y+1=0$ and it's length $=\sqrt{2}$ units
Ex 9.3 Question 18.

If $p$ is the length of perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$, then show that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.

Answer.

Given: Line $\frac{x}{a}+\frac{y}{b}=1$
$
\begin{aligned}
& \Rightarrow b x+a y=a b \\
& \Rightarrow b x+a y-a b=0
\end{aligned}
$

Now, $p$ is the length of perpendicular from origin to $b x+a y-a b=0$.
We have the length of the perpendicular from of a given point $\left(x_1, y_1\right)$ to a given line $\mathrm{Ax}+\mathrm{BY}+\mathrm{C}=0$ is $d=\left|\frac{A x_1+B y_1+C}{\sqrt{A^2+B^2}}\right|$
$
\therefore p=\left|\frac{b \times 0+a \times 0-a b}{\sqrt{b^2+a^2}}\right|=\frac{a b}{\sqrt{b^2+a^2}}
$

$\begin{aligned}
& \Rightarrow p^2=\frac{a^2 b^2}{b^2+a^2} \\
& \Rightarrow \frac{1}{p^2}=\frac{b^2+a^2}{a^2 b^2} \\
& \Rightarrow \frac{1}{p^2}=\frac{b^2}{a^2 b^2}+\frac{a^2}{a^2 b^2} \\
& \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}
\end{aligned}$