Finding the coordinate C. A triangle $A$, $B$, $C$ has the coordinates:
$A = (-1, 3)$
$B = (3, 1)$
$C = (x, y)$
$BC$ is perpendicular to $AB$. Find the coordinates of $C$

My attempt:
Grad of $AB$ =
$$\frac{3-1}{-1-3} = -0.5$$
Grad of $BC = 2$ ($-0.5 \times 2 = -1$ because AB and BC are perpendicular).
Equation of $BC$
$(y-1) = 2(x-3)$
$y = 2x - 5$
Equation of $AC$
$(y-3) = m(x--1)$
$y = mx+m+3$
I do not know how to proceed further. Please help me out.

The equation of $AC$ is $x-3y=-10$ as slope of any line is $-\frac{a}{b}$ where $a$ is $x$-coordinate and $b$ is $y$-coordinate so slope is $-\left(\frac{-1}{3}\right)$. $C$ is the point where $AC$ and $BC$ so meet we have two simultaneous equations $2x-y=5$ and $x-3y=-10$ solving them you get $x=5$ and $y=5$.  
