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.

Yes, the gradient of line AB is $-1/2$ so the gradient of any line perpendicular to $AB$ is $2$. The line with gradient $2$ through $B = (3, 1)$ is $y = 2(x- 3)+ 1$ or $y= 2x- 5$.  All points on that line, $(x, 2x- 5)$, satisfy the conditions for C.
