Zauważmy, że aby trójkąt był równoramienny odległość punktu A od prostej k musi być taka sama jak odległość punktu B od prostej k. Zapiszmy

$2x-3y+19=0\mid +3y$  

$3y=2x+19\mid :3$  

$y=\frac{2}{3}x+\frac{19}{3}$  

$C=\left(x_C,\frac{2}{3}{x}_C+\frac{19}{3}\right)$  

Zatem mamy równanie

$\left|AC\right|=\sqrt{{\left(5-x_C\right)}^2+{\left(2-\frac{2}{3}{x}_C-\frac{19}{3}\right)}^2}=\sqrt{{\left(-1-x_C\right)}^2+{\left(4-\frac{2}{3}{x}_C-\frac{19}{3}\right)}^2}=\left|BC\right|$  

$\sqrt{{\left(5-x_C\right)}^2+{\left(2-\frac{2}{3}{x}_C-\frac{19}{3}\right)}^2}=\sqrt{{\left(-1-x_C\right)}^2+{\left(4-\frac{2}{3}{x}_C-\frac{19}{3}\right)}^2}{\mid }^2$  

${\left(5-x_C\right)}^2+{\left(2-\frac{2}{3}{x}_C-\frac{19}{3}\right)}^2={\left(-1-x_C\right)}^2+{\left(4-\frac{2}{3}{x}_C-\frac{19}{3}\right)}^2$  

$25-10x_C+\cancel{x_C^2}+{\left(-\frac{2}{3}{x}_C-\frac{13}{3}\right)}^2=1+2x_C+\cancel{x_C^2}+{\left(-\frac{2}{3}{x}_C-\frac{7}{3}\right)}^2$  

$25-10x_C+\frac{4}{9}{x}_C^2+2\cdot \frac{2}{3}\cdot \frac{13}{3}{x}_C+\frac{169}{9}=1+2x_C+\frac{4}{9}{x}_C^2+2\cdot \frac{2}{3}\cdot \frac{7}{3}{x}_C+\frac{49}{9}$  

$25-10x_C+\cancel{\frac{4}{9}{x}_C^2}+\frac{52}{9}{x}_C+\frac{169}{9}=1+2x_C+\cancel{\frac{4}{9}{x}_C^2}+\frac{28}{9}{x}_C+\frac{49}{9}$  

$25-10x_C+\frac{52}{9}{x}_C+\frac{169}{9}=1+2x_C+\frac{28}{9}{x}_C+\frac{49}{9}\mid \cdot 9$  

$225-90x_C+52x_C+169=9+18x_C+28x_C+49$  

$394-38x_C=58+46x_C\mid +38x_C-58$  

$84x_C=336\mid :84$  

$x_C=4$  

$y_C=\frac{2}{3}{x}_C+\frac{19}{3}=\frac{2}{3}\cdot 4+\frac{19}{3}=\frac{8}{3}+\frac{19}{3}=\frac{27}{3}=9$  

$C=\left(4,9\right)$