Compute the intersection of perpendicular lines

Question

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:

$$ Ax+By-C= Bx-Ay $$ $$By+Ay =-Ax +C +Bx $$ $$y(B+A) = -Ax +C +Bx$$ $$y= \frac{-Ax + Bx +C}{ B +A} $$

$$y= \frac{x(-A +B) +C}{ B +A} $$

Answer 1

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$$y=\frac{BC}{A^2+B^2}.$$

Answer 2

If $B \ne 0$, then

$Bx - Ay = 0 \iff x = \dfrac ABy$

Substituting into $Ax+By=C$, we get

\begin{align} \dfrac{A^2}{B}y+By=C &\implies A^2y + B^2y = BC \\ &\implies y = \dfrac{BC}{A^2+B^2} \\ &\implies x = \dfrac ABy = \dfrac{AC}{A^2+B^2} \end{align}

If $B=0$ and $A \ne 0$, then $Ax+By=C \implies Ax=C \implies x = \dfrac CA$ and $Bx-Ay=0 \implies -Ay=0 \implies y=0$. Which is consistent with

$$(x,y) = \left(\dfrac{AC}{A^2+B^2}, \dfrac{BC}{A^2+B^2} \right)$$

Answer 3

Using Cramer's rule: $$x=\frac{|A_x|}{|A|}=\frac{\begin{vmatrix}C & B \\ 0 & -A\end{vmatrix}}{\begin{vmatrix}A & B \\ B & -A\end{vmatrix}}=\frac{AC}{A^2+B^2}.$$ You try to show for $y$.