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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

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4x + 5y = 20..........(i)
3x - 2y + 6 = 0..........(ii)
Orthocenter (1, 1)
Line to 4x + 5y = 20 and passes (1, 1) is 5x - 4y = 1..........(iii)
and line to 3x - 2y + 6 = 0 and passes (1, 1) is 2x + 3y = 5..........(iv)
on solving (i) & (iv) pt (352,−10)\left( \frac{35}{2}, - 10 \right)
on solving (ii) & (iii) pt (−13,−332)\left( - 13, - \frac{33}{2} \right)
side BC is y + 10 =
−332+10−13−352(x−352)\frac{- \frac{33}{2} + 10}{- 13 - \frac{35}{2}}\left( x - \frac{35}{2} \right)
⇒y+10=1361(x−352)\Rightarrow y + 10 = \frac{13}{61}\left( x - \frac{35}{2} \right)
⇒26x−122y−1675=0\Rightarrow 26x - 122y - 1675 = 0

Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx

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