The point $(a, b)$ is the foot of the perpendicular drawn from the point $(3,1)$ to the line $x+3 y+4=0$. If…
The point $(a, b)$ is the foot of the perpendicular drawn from the point $(3,1)$ to the line $x+3 y+4=0$. If $(p, q)$ is the image of $(a, b)$ with respect to the line $3 x-4 y+11=$ 0 , then $\frac{p}{a}+\frac{q}{b}=$
$-3$
$-5$
$3$
$7$
Solution
Since, $(a, b)$ is the foot of perpendicular drawn from $(3,1)$ to the line $x+3 y+4=0$.
$\Rightarrow a+3 b+4=0...(i)$
Now, equation of a line passing through $(3,1)$ and perpendicular to the given line is $3 x-y-8=0$
so $3 a-b-8=0$...(ii)
After solving (i) and (ii), we get $(a, b)=(2,-2)$
Let $P=\left(\frac{2+p}{2}, \frac{-2+q}{2}\right)$
Since, $P$ lies on $3 x-4 y+11=0$
$\Rightarrow \frac{3(2+p)}{2}-\frac{4(-2+q)}{2}+11=0$
$\Rightarrow 3 p-4 q+36=0...(iii)$
Since slope of line $3 x-4 y+11=0$ is $\frac{3}{4}$
Now, $\frac{3}{4} \times\left(\frac{-2-q}{2-p}\right)=-1 \Rightarrow \frac{2+q}{2-p}=\frac{4}{3}$
$\Rightarrow 4 p+3 q-2=0....(iv)$
After solving (iii) and (iv) we get, $p=-4, q=6$ Now, $\frac{p}{a}+\frac{q}{b}=\frac{-4}{2}+\frac{6}{-2}=-2-3=-5$.