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 $3x-4y+25=0$.

Answer 1

Answer: 10

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 $3x-4y+25=0$
$\text{ is }=\frac{|3(3)-4(-4)+25|}{\sqrt{(3{)}^2+(-4{)}^2}}=\frac{50}{5}=10$