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  4. 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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Q. 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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Solution:

Equation of $AB$ is $3x - 2y + 6 = 0$
Equation of $AC$ is $4x + 5y - 20 = 0$
Equation of $BE$ is $2x + 3y - 5 = 0$
Equation of $CF$ is $5x - 4y - 1 = 0$
$\Rightarrow $ Equation of BC is $26x - 122y = 1675$

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