2015/070) Find the point on the line $2 x + 4 y + 1 = 0$ which is closest to the point $(-1, -3 )$

There are different methods to solve this problem. I shall use one of them.
 
Because the point is closest to $(-1,-3)$ the line joining the point and $(-1,-3)$ shall be perpendicular to $(2x + 4y + 1) = 0$

slope of $2x + 4y + 1 = 0$ is $\dfrac{- 1}{2}$
so slope of perpendicular line is 2
so the equation of line
$y = 2x + c$
as it passes through $(-1,-3)$
so $-3 = - 2 + c$ or  $c = -1$
so $y = 2x -1$
solving this and $2x + 4y + 1 = 0$ we get the result
$2x + 4 ( 2x-1) + 1 = 0$
or $10x - 3 = 0$
$x = \dfrac{3}{10}$ and $y = \dfrac{- 2}{5}$

So  the point is $(\dfrac{3}{10}, \dfrac{- 2}{5})$