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  4. The orthocentre of triangle formed by lines 4 x-7 y+10=0, x+y=5 and 7 x+4 y=15 is

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Q. The orthocentre of triangle formed by lines $4 x-7 y+10=0$, $x+y=5$ and $7 x+4 y=15$ is

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Solution:

Orthocentre of right angled triangle is same as the vertex of right angle. Therefore, point of intersection of $7 x$ $+4 y-15=0$ and $4 x-7 y+10=0$ is $(1,2)$