Como se obtém o pé da perpendicular de um ponto P(x0,y0) sobre a reta ax+by+c=0?

x' e y' dados por \(x'=x_0-\dfrac{a(ax_0+by_0+c)}{a^2+b^2},\; y'=y_0-\dfrac{b(ax_0+by_0+c)}{a^2+b^2}\).

Exemplo:
Considere P(1,2) e a reta \(2x-y+1=0\) (\(a=2,b=-1,c=1\)).
Calcule \(D=ax_0+by_0+c=2\cdot1+(-1)\cdot2+1=1\).
\(a^2+b^2=4+1=5\).
Então \(x'=1-\dfrac{2\cdot1}{5}=\dfrac{3}{5},\; y'=2-\dfrac{(-1)\cdot1}{5}=\dfrac{11}{5}\).
O pé é \(M\!\left(\tfrac{3}{5},\tfrac{11}{5}\right)\).

Escolhendo \(Q(4,0)\) e \(R(5,3)\), o ponto médio de \(QR\) é \(N\!\left(\tfrac{9}{2},\tfrac{3}{2}\right)\).
Área do triângulo \(PMN\) por determinante:
\(\displaystyle [PMN]=\tfrac{1}{2}\big|x_P(y_M-y_N)+x_M(y_N-y_P)+x_N(y_P-y_M)\big|\)
\(\displaystyle =\tfrac{1}{2}\Big|1\big(\tfrac{11}{5}-\tfrac{3}{2}\big)+\tfrac{3}{5}\big(\tfrac{3}{2}-2\big)+\tfrac{9}{2}\big(2-\tfrac{11}{5}\big)\Big|\)\
\(\displaystyle =\tfrac{1}{2}\big|\tfrac{7}{10}-\tfrac{3}{10}-\tfrac{9}{10}\big|=\tfrac{1}{2}\cdot\tfrac{1}{2}=\tfrac{1}{4}=0{,}25\).\