Question

Let the point $M (2,1)$ be shifted through a distance $3 \sqrt{2}$ units measured parallel to the line $L : x + y -1=0$ in the direction of decreasing ordinates, to reach at $N$. If the image of $N$ in the line $L$ is $R$, then find the distance of $R$ from the line $3 x-4 y+25=0$.

Answer 1

We have $MN =3 \sqrt{2}$

$\therefore x _{ N }=2-3 \sqrt{2}\left(\frac{-1}{\sqrt{2}}\right)=5 $
$\text { and } y _{ N }=1-3 \sqrt{2}\left(\frac{1}{\sqrt{2}}\right)=-2 $
$\text { so, } N (5,-2)$
Image of $N$ in $x + y -1=0$ is $(3,-4)$
$\therefore R (3,-4)$
So, distance of $R (3,-4)$ from the line $3 x-4 y+25=0$
$\text { is }=\frac{|3(3)-4(-4)+25|}{\sqrt{(3)^2+(-4)^2}}=\frac{50}{5}=10 $