The orthocentre of the triangle formed by the lines x + 3y – 10 = 0 and $6 x^{2} + x y - y^{2} = 0 i s$

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(1, -3)

(-1, 3)

(3, 1)

(1, 3)

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 6 x^{2} + x y - y^{2} = 0 \\ C o m p a r i n g w i t h a x^{2} + 2 h x y + b y^{2} = 0 \\ a = 6, 2 h = 1, b = - 1 \\ a n d x + 3 y - 10 = 0 w i t h l x + m y + n = 0 \\ l = 1, m = 3, n = - 10 \\ k = \frac{- n (a + b)}{a m^{2} - 2 h l m + b l^{2}} \end{array}$

$\begin{array}{l} = \frac{10 (6 - 1)}{6 (9) - 1 (1) (3) + (- 1) 1} \Rightarrow \frac{10 (5)}{54 - 3 - 1} = \frac{50}{50} = 1 \\ ∴ O r t h o c e n t r e = (k l, k m) = (1 (1), 1 (3)) \Rightarrow (1, 3) \end{array}$