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.

HINT: 
Length of $AC =$
$\sqrt{((x+1)^2 +(y-3)^2)} = \sqrt{(\mathrm{Length} \space \mathrm{of} \space (AB)^2+\mathrm{Length} \space \mathrm{of} \space (BC)^2)}\tag{1}$ 
because you have a right-angled triangle as $BC$ is perpendicular to $AB$
From $(1)$ $$(x+1)^2 +(y-3)^2=(3+1)^2+(1-3)^2+(x-3)^2+(y-1)^2$$
$$\implies x^2+2x+1 +y^2-6y+9=20+x^2-6x+9+y^2-2y+1$$
