Exercise 6.3-Additional Questions - Chapter 6 - Two Dimensional Analytical Geometry - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions
Question 1.
Find the equation of the line passing through the point $(5,2)$ and perpendicular to the line joining the points $(2,3)$ and $(3,-1)$.
Solution:
Slope of the line joining the points $(2,3)$ and $(3,-1)$ is
$
\frac{-1-3}{3-2}=-4
$
Slope of the required line which is perpendicular to it
$
=\frac{-1}{-4}=\frac{1}{4} \quad\left[\because m_1 m_2=-1\right]
$
Equation of the line passing through the point $(5,2)$ is
$
\begin{array}{rlrl}
y-2 & =\frac{1}{4}(x-5) & {\left[y-y_1=m\left(x-x_1\right)\right]} \\
\Rightarrow & & 4 y-8 & =x-5 \\
\Rightarrow & x-4 y+3 & =0
\end{array}
$
Hence, the required equation is $x-4 y+3=0$.

Question 2.
Find the points on the line $x+y=4$ which lie at a unit distance from the line $4 x+3 y=10$.

Solution

 Let $\left(x_1, y_1\right)$ be any point lying in the equation $x+y=4$
$
\therefore \quad x_1+y_1=4
$
Distance of the point $\left(x_1, y_1\right)$ from the equation $4 x+3 y=10$
$
\begin{aligned}
\frac{4 x_1+3 y_1-10}{\sqrt{(4)^2+(3)^2}} & =1 \\
\left|\frac{4 x_1+3 y_1-10}{5}\right| & =1 \\
4 x_1+3 y_1-10 & = \pm 5
\end{aligned}
$
Taking $(+) \operatorname{sign} \quad 4 x_1+3 y_1-10=5$
$
\Rightarrow \quad 4 x_1+3 y_1=15
$
From equation $(i)$ we get $y_1=4-x_1$
Putting the value of $y_1$ in equation (ii) we get
$
\begin{array}{rlrl} 
& & 4 x_1+3\left(4-x_1\right) & =15 \\
\Rightarrow & 4 x_1+12-3 x_1 & =15 \\
\Rightarrow & x_1+12 & =15 \\
\Rightarrow & x_1 & =3 \text { and } y_1=4-3=1
\end{array}
$
So, the required point is $(3,1)$
Now taking(-) sign, we have
$
\begin{aligned}
& 4 x_1+3 y_1-10=-5 \\
& \Rightarrow \quad 4 x_1+3 y_1=5 \\
& \Rightarrow \quad 4 x_1+3\left(4-x_1\right)=5 \\
& \Rightarrow \quad 4 x_1+12-3 x_1=5 \\
& \Rightarrow \quad x_1=5-12=-7 \\
& \text { and } y_1=4-(-7)=11 \\
&
\end{aligned}
$
So, the required point is $(-7,11)$
Hence, the required points on the given line are $(3,1)$ and $(-7,11)$.

Question 3 .
Find the equation of the line passing through the point of intersection $2 x+y=5$ and $x+3 y+8=0$ and parallel to the line $3 x+4 y=7$.
Solution:
Given that:
$
\begin{array}{r}
2 x+y=5 \\
x+3 y+8=0 \\
3 x+4 y=7
\end{array}
$
Equation of any line passing through the point of intersection of equation $(i)$ and (ii) is
$
\begin{aligned}
(2 x+y-5)+\lambda(x+3 y+8) & =0 \\
\Rightarrow \quad 2 x+y-5+\lambda x+3 \lambda y+8 \lambda & =0 \\
\Rightarrow \quad(2+\lambda) x+(1+3 \lambda) y-5+8 \lambda & =0 \\
\quad \text { Slope of line } m_1 \text { (say) } & =\frac{-(2+\lambda)}{1+3 \lambda} \quad\left[\because m=\frac{-a}{b}\right]
\end{aligned}
$
Now slope of line $3 x+4 y=7$ is
$
m_2 \text { (say) }=-\frac{3}{4}
$
If equation (iii) is parallel to equation (iv) then $m_1=m_2$
$
\begin{array}{ll}
\Rightarrow & \frac{-(2+\lambda)}{1+3 \lambda}=-\frac{3}{4} \\
\Rightarrow & \frac{2+\lambda}{1+3 \lambda}=\frac{3}{4} \Rightarrow 8+4 \lambda=3+9 \lambda \\
\Rightarrow & 9 \lambda-4 \lambda=5 \Rightarrow 5 \lambda=5 \Rightarrow \lambda=1
\end{array}
$
On putting the value of $\lambda$ in equation (iv) we get
$
\begin{aligned}
& (2 \mathrm{x}+\mathrm{y}-5)+1(\mathrm{x}+3 \mathrm{y}+8)=0 \\
& \Rightarrow 2 \mathrm{x}+\mathrm{y}-5+\mathrm{x}+3 \mathrm{y}+8=0
\end{aligned}
$

$
\Rightarrow 3 x+4 y+3=0
$
Hence, the required equation is $3 x+4 y+3=0$
Question 4.
A line passing through the points (a, 2a) and $(-2,3)$ is perpendicular to the line $4 x+3 y+5=0$, find the value of a.
Solution:
Let $m_1$ be the slope of the line joining A $(a, 2 a)$ and $\mathrm{B}(-2,3)$. Then $m_1=\frac{2 a-3}{a+2}$
Let $m_2$ be the slope of the line $4 x+3 y+5=0$. Then $m_2=-\frac{4}{3}$ Since given lines are perpendicular. Therefore,
$
m_1 m_2=-1 \Rightarrow \frac{2 a-3}{a+2} \times-\frac{4}{3}=-1 \Rightarrow 8 a-12=3 a+6 \Rightarrow a=18 / 5
$

Question 5.
Find the equation of the straight line which passes through the intersection of the straight lines $2 x+y=8$ and $3 x-2 y+7=0$ and is parallel to the straight line $4 x+y-11=0$.
Solution:
Equation of line through the intersection of straight lines
$2 x+y=8$ and $3 x-2 y+7=0$ is
$2 \mathrm{x}+\mathrm{y}-8+\mathrm{k}(3 \mathrm{x}-2 \mathrm{y}+7)=0$
$\mathrm{x}(2+3 \mathrm{k})+\mathrm{y}(1-2 \mathrm{k})+(-8+7 \mathrm{k})=0$
Then $m_1=-\left(\frac{2+3 k}{1-2 k}\right)$
This is parallel to $4 x+y-11=0$
Then $m_2=-\left(\frac{4}{1}\right)$
$\therefore$ Their slopes are equal $\Rightarrow m_1=m_2$
$
-\left(\frac{2+3 k}{1-2 k}\right)=-\left(\frac{4}{1}\right)
$
$
\begin{aligned}
\frac{2+3 k}{1-2 k} & =4 \\
2+3 k & =4-8 k \\
11 k & =2 \Rightarrow k=\frac{2}{11}
\end{aligned}
$
Required equation is $x\left(2+\frac{6}{11}\right)+y\left(1-\frac{4}{11}\right)+\left(-8+\frac{14}{11}\right)=0$
$
\begin{aligned}
& \frac{28 x}{11}+\frac{7 y}{11}-\frac{74}{11}=0 \\
& \Rightarrow 28 x+7 y-74=0 \\
& \Rightarrow 28 x+7 y-74=0
\end{aligned}
$

Question 6

Find the equation of the straight line passing through intersection of the straight lines $5 x-6 y=1$ and $3 x$ $+2 y+5=0$ and perpendicular to the straight line $3 x-5 y+11=0$.
Solution:
Equation of line through the intersection of straight lines $5 x-6 y=1$ and $3 x+2 y+5=0$ is $5 \mathrm{x}-6 \mathrm{y}-1+\mathrm{k}(3 \mathrm{x}+2 \mathrm{y}+5)=0$
$\mathrm{x}(5+3 \mathrm{k})+\mathrm{y}(-6+2 \mathrm{k})+(-1+5 \mathrm{k})=0$
This is perpendicular to $3 x-5 y+11=0$
That is, the product of their slopes is -1
$
\begin{aligned}
-\left(\frac{5+3 k}{-6+2 k}\right)\left(-\frac{3}{-5}\right) & =-1 \\
\Rightarrow \frac{15+9 k}{-30+10 k} & =1 \\
\Rightarrow 15+9 k & =-30+10 k \\
45 & =k
\end{aligned}
$
Required equation is $5 x-6 y-1+45(3 x+2 y+5)=0$
$
\begin{aligned}
5 x-6 y-1+135 x+90 y+225 & =0 \\
140 x+84 y+224 & =0 \\
20 x+12 y+32 & =0 \\
5 x+3 y+8 & =0
\end{aligned}
$