Skoro trójkąt jest równoramienny, to $\left|AC\right|=\left|BC\right|$.
Mamy: $A=\left(-10,5\right)$, $B=\left(2,-3\right)$, $C=\left(x,\frac{19}{4}x-\frac{25}{2}\right)$

$\sqrt{{\left(x-\left(-10\right)\right)}^2+{\left(\frac{19}{4}x-\frac{25}{2}-5\right)}^2}=\sqrt{{\left(x-2\right)}^2+{\left(\frac{19}{4}x-\frac{25}{2}-\left(-3\right)\right)}^2}$

${\left(x+10\right)}^2+{\left(\frac{19}{4}x-\frac{25}{2}-\frac{10}{2}\right)}^2={\left(x-2\right)}^2+{\left(\frac{19}{4}x-\frac{25}{2}+\frac{6}{2}\right)}^2$

${\left(x+10\right)}^2+{\left(\frac{19}{4}x-\frac{35}{2}\right)}^2={\left(x-2\right)}^2+{\left(\frac{19}{4}x-\frac{19}{2}\right)}^2$

$x^2+20x+100+\frac{361}{4}{x}^2-2\cdot \frac{19}{4}x\cdot \frac{35}{2}+\frac{1225}{4}=x^2-4x+4+\frac{361}{4}{x}^2-2\cdot \frac{19}{4}x\cdot \frac{19}{2}+\frac{361}{4}$ 

$x^2+20x+100+\frac{361}{4}{x}^2-\frac{665}{4}x+\frac{1225}{4}=x^2-4x+4+\frac{361}{4}{x}^2-\frac{361}{4}x+\frac{361}{4}|-x^2-\frac{361}{4}{x}^2$

$20x+100-\frac{665}{4}x+\frac{1225}{4}=-4x+4-\frac{361}{4}x+\frac{361}{4}|\cdot 4$

$80x+400-665x+1225=-16x+16-361x+361$