Let the equations 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

Let the equation of side AB $3x-2y+6=0$  

                                                                                                          …(i)

and, the equation of side AC is 4x + 5y - 20 = 0

                                                                                                          …(ii)

Solving (i) and (ii), we get vertex  $A\left(\frac{10}{23},\frac{84}{23}\right)$.

Now, slope of $AC=-4/5$

$\therefore$ Slope of perpendicular $BE=5/4$

Also, BE passes through orthocentre H(l, 1)

$\therefore$ Equation of BE is

$(y-1)=\frac{5}{4}(x-1)$

$\Rightarrow 4y-5x+1=0$                                                      …(iii)

Solving (i) and (iii), we get vertex $B\left(-13,-\frac{33}{2}\right)$.

Now, slope of $AB=3/2$

$\therefore$ Slope of perpendicular $CF=-2/3$

Equation of CF is $(y-1)=\frac{-2}{3}(x-1)$

$\Rightarrow 3y+2x-5=0$                                                          ….(iv)                             

Solving (ii) and (iv), we get vertex $C\left(\frac{35}{2},-10\right)$

Equation of side BC is  $(y+10)=\frac{-10+\frac{33}{2}}{\frac{35}{2}+13}\left(x-\frac{35}{2}\right)$

$\Rightarrow (y+10)=\frac{13}{61}\left(x-\frac{35}{2}\right)\Rightarrow 26x-122y=1675$