Question

Find the ortho-centre of the triangle formed by the lines $3x-4y=10,8x+6y=15$ and y-axis.

Answer 1

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$3x-4y=10$ .......(1)

$8x+6y=15$ ......(2)

$4x+3y=\frac{15}{2}$

Equation 1 & 2 are perpendicular lines their intersecting point is ortho-centre.

On solving eqaution 1 & 2 , we get the ortho-centre as $(\frac{12}{5},\frac{-7}{10})$.

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