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If A is the orthocentre of triangle formed by $2 x^{2} - y^{2} = 0,$ $x + y - 1 = 0$ and B is the centroid of the triangle formed by $2 x^{2} - 5 x y + 2 y^{2} = 0,$ $7 x - 2 y - 12 = 0$ then the distance between A and B is

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$\sqrt{5}$

$1$

$5$

$\sqrt{2}$

answer is A.

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Detailed Solution

Centriod $= (2, 3)$

$\begin{aligned} a = 2, h = 0, b = - 1 and l = 1, m = 1, n = - 1 \\ K = \frac{- n (a + b)}{a m^{2} - 2 h l m + b l^{2}} = \frac{1 (2 - 1)}{2 (1) - 0 - 1} = 1 \end{aligned}$

$H = orthocentre = (k l, k m) = (1, 1)$
Ortho centre $= (1, 1)$ (from synopsis )
$H G = \sqrt{(2 - 1)^{2} + (3 - 1)^{2}} = \sqrt{1 + 4} = \sqrt{5}$

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