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)=54(x−1)⇒5x−4y=1.......(3)and line perpendicular to 3x-2y+6=0 and passes through (1,1)y−1=−23(x−1)⇒2x+3y=5........(4)Solving (1) and (4) we get C352,−10Solving (2) and (3) we get A−13,−332Equation of BC is 26x-122y-1675=0