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