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The orthocenter of the triangle formed by the lines $x + 3 y = 10$ and $6 x^{2} + x y - y^{2} = 0$ is

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$(1, - 3)$

$(1, 3)$

$(- 1, 3)$

$(3, 1)$

answer is A.

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Detailed Solution

$6 x^{2} + x y - y^{2} = 0 \Rightarrow (2 x + y) (3 x - y) = 0 \Rightarrow 2 x + y = 0, 3 x - y = 0$

$∴ x + 3 y = 10, 2 x + y = 0, 3 x - y = 0$
Form a right angled triangle orthocenter = point of intersection of $(x + 3 y = 10, 3 x - y = 0) = (1, 3)$

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