Equation of two sides of a triangle are 3x-2y+6=0 and 4x-5y-20=0 and the orthocenter is (1,1). Find the equation of the third side.

Question

Equation of two sides of a triangle are 3x-2y+6=0 and 4x-5y-20=0
and the orthocenter is (1,1) Find the equation of the third
side

Vipin Verma · Accepted Answer

Here BD and CF are perpendicular to each other
And they intersect at O

For a line to be perpendicular to a line just change the
coefficient of x and y , with either x or y with minus sign
So Equation of BD = 4y + 5x +C = 0, here C is some constant
As BD is passing through (1,1), so C = -9
So equation of BD = 4y + 5x -9 =0

And Equation of CF = 2x + 3y + C1 =0
As this is also passing through (1,1), so C1 = -5
Hence the equation of CF = 2x +3y -5 =0

As BD will intersect AB at B, so solving BD and AB, we get
B (-3/11, 57/22 )

And CF intersect AC at C, so on solving both equation we get
C as (85/22 , -10/11)

So as we found B and C, so the equation of BC :
(y - 57/22) = {(57/22 + 10/11)/(-3/11 - 85/22)} ( x + 3/11)