Compute the intersection of perpendicular lines Lines Ax + By = C and Bx – Ay = 0 are perpendicular.
I am trying to understand how their intersection is computed below:
$$ x = \frac{AC}{A^2 + B^2}$$
$$ y= \frac{BC}{A^2 + B^2}$$
I tried solving for y first but I am  not sure how to continue:
$$ Ax+By-C= Bx-Ay $$
$$By+Ay =-Ax +C +Bx $$
$$y(B+A) = -Ax +C +Bx$$
$$y= \frac{-Ax + Bx +C}{ B +A} $$
$$y= \frac{x(-A +B) +C}{ B +A} $$

Assuming that $(A,B)\neq(0,0)$:\begin{align}\left\{\begin{array}{l}Ax+By=C\\Bx-Ay=0\end{array}\right.&\implies A(Ax+BY)+B(Bx-Ay)=AC\\&\iff(A^2+B^2)x=AC\\&\iff x=\frac{AC}{A^2+B^2}.\end{align}From this and from $Bx=Ay$, you get that$$y=\frac{BC}{A^2+B^2}.$$
