Let the equation of two sides of a triangle be $3x-2y+6=0$ and $4x+5y-20=0$. If the orthocentre of this triangle is at (1,1). then the equation of its third side is

$4x+5y-20=0$………(1)

$3x-2y+6=0$…….. (2)

orthocentre is (1,1)

lien perpendicular to $4x+5y-20=0$ and passes through (1,1) is $(y-1)=\frac{5}{4}(x-1)$

$\Rightarrow 5x-4y=\mathrm{1.......}\left(3\right)$

and line perpendicular to $3x-2y+6=0$ and passes through (1,1)

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

$\Rightarrow 2x+3y=\mathrm{5........}\left(4\right)$

Solving (1) and (4) we get $C\left(\frac{{\displaystyle 35}}{{\displaystyle 2}},-10\right)$

Solving (2) and (3) we get $A\left(-13,\frac{{\displaystyle -33}}{{\displaystyle 2}}\right)$

Equation of BC is $26x-122y-1675=0$