Let the equations of two sides of a triangle be 3x - 2y + 6 = 0 and...

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

${m}_{AC} = \dfrac{3}{2}$

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

hence ${m}_{B{B}^{'}} = - \dfrac{2}{3}$

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 $B{B}^{'}$

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

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

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

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

Equation of $BC$ is, $y - {y}_{1} = \dfrac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}\left(x - {x}_{1}\right)$

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

$∴ 26x - 122y - 1675 = 0$