Let the equations of two sides of a triangle be $3x-2y+6=0$ and $4x+5y-20=0.$ If the orthocenter of this triangle is at $\left(1, 1\right)$ then the equation of it's third side is:

To find equation of $BC,$ first we will find coordinates of $B$ and $C.$

$m_{AC}=\frac{3}{2}$

$AC$ and $B{B}^{\text{'}}$ are perpendicular to each other 

hence $m_{B{B}^{\text{'}}}=-\frac{2}{3}$

Equation of line passing through $\left(x_1,y_1\right)$ and having slope $m$ is $y-y_1=m\left(x-x_1\right)$

For equation of $B{B}^{\text{'}}$

$y-1=\frac{-2}{3}\left(x-1\right)$

$\Rightarrow 2x+3y=5 ........\left(1\right)$ 

Point of intersection of $AB$ and $B{B}^{\text{'}}$ $\Rightarrow B \left(\frac{35}{2}, -10\right)$

Similarly point $C\left(-13, -\frac{33}{2}\right)$

Equation of $BC$ is, $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

$y+10=\frac{13}{61} \left(x-\frac{35}{2}\right)$

$\therefore 26x-122y-1675=0$