If the straight lines $2 x + 3 y - 1 = 0, x + 2 y - 1 = 0$ and $a x + b y - 1 = 0$ form a triangle with origin as ortho centre, then $(a, b) = ?$

(6, 4)

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(- 3, 3)

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(8, - 8)

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(0, 7)

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Solution

The correct option is D $(8, - 8)$
Intersection of $2 x + 3 y - 1 = 0, x + 2 y - 1 = 0$ is $(- 1, 1) .$
Line passing through $(- 1, 1)$ and origin is perpendicular to $a x + b y - 1 = 0,$
we get $a = - b .$ ---1) using $m_{1} . m_{2} = - 1$
Also, line passing through $x + 2 y - 1 = 0$ and $a x + b y - 1 = 0,$ is
$x + 2 y - 1 + λ (a x + b y - 1) = 0,$
It passes through origin , so
$λ = - 1$
Line is $x (1 - a) + y (2 - b) = 0$
This is perpendicular to line $2 x + 3 y - 1 = 0$
So, using $m_{1} . m_{2} = - 1$ we get
$2 a + 3 b = 8$ ---2)
Then, on solving further, we get $(a, b) = (8, - 8)$

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