The orthocenter of the triangle formed by the lines $x-2 y=10$ and $6 x^2+x y-y^2=0$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 6 x^2+x y-y^2=0 \ & \Rightarrow(2 x+y)(3 x-y)=0 \ & \Rightarrow 3 x-y=0 	ext { and } 2 x+y=0 	ext { and } x-2 y=10 \ & \because 2 x+y=0 	ext { and } x-2 y=10 	ext { are perpendicular }\end{aligned}$ Hence, orthocentre is point of intersection $2 x+y=0$ and $x-2 y=10$ i.e. $(2,-4)$

The orthocenter of the triangle formed by the lines $x-2 y=10$ and $6 x^2+x y-y^2=0$ is

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