Official solution
4x + 5y - 20 = 0 . . .(1)
3x - 2y + 6 = 0 . . . (2)
orthocentre is (1, 1)
line perpendicular to 4x + 5y - 20 = 0
and passes through (1, 1) is
(y - 1) = \[{5 \over 4}(x - 1)
\[ \Rightarrow \] 5x - 4y = 1 . . .(3)
and line \[ \bot \] to 3x - 2y + 6 = 0
and passes through (1, 1)
y - 1 = - \[{2 \over 3}\] (x - 1)
\[ \Rightarrow \] 2x + 3y = 5 . . .(4)
Solving (1) and (4) we get C\[\left( {{{35} \over 2}, - 10} \right)\]
Solving (2) and (3) we get A \[\left( { - 13,{{ - 33} \over 2}} \right)\]
Side BC is y + 10 = \[{{{{ - 33} \over 2} + 10} \over { - 13 - {{35} \over 2}}}\left( {x - {{35} \over 2}} \right)\]
\[ \Rightarrow \] y + 10 = \[{{13} \over {61}}\left( {x - {{35} \over 2}} \right)\]
\[ \Rightarrow \] 26x - 122y - 1675 = 0