Q. The image of the point $(2,3)$ on the line $x+3 y+4=0$ is

Solution:

Using fact (short cut) the image of $(h, k)$ in the line $a x+b y+c=0$ is given by
$\frac{x-h}{a}=\frac{y-k}{b}=-\frac{2(a h+b k+c)}{a^{2}+b^{2}}$
$ \therefore \frac{x-2}{1}=\frac{y-3}{3}=-\frac{2(1)+3(3)+4}{10} $
$ x =-1, y =-6$