The lines joining the joining the origin to the points of intersection of $x + 2 y = k$ with the pair of lines $2 x^{2} - 2 x y + 3 y^{2} + 2 x - y - 1 = 0$ are at right angles then k=

\pm 1

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\pm 2

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\pm 3

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\pm 4

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Solution

The correct option is A $\pm 1$
We use homogenisation to find $O A \times O B$ where $A$ and $B$ are intersecpts of line on pair of lines and $O$ is the origin.
$\Rightarrow 2 x^{2} + 3 y^{2} - 2 x y + (2 x - y) (\frac{x + 2 y}{k}) - (\frac{x + 2 y}{k})^{2} = 0$
$\Rightarrow 2 k^{2} x^{2} + 3 k^{2} y^{2} - 2 k^{2} x y + 2 k x^{2} + 4 k x y - k x y - 2 k y^{2} - (x^{2} + 4 y^{2} + 4 x y) = 0$
$\Rightarrow x^{2} (2 k^{2} + 2 k - 1) + y^{2} (3 k^{2} - 2 k - 4) + x y (4 k - 2 k^{2} - k - 4) = 0$
$\Rightarrow O A$ and $O B$ are perpendicular.
$\Rightarrow m_{1} m_{2} = \frac{a}{b} = - 1$
$\Rightarrow 2 k^{2} + 2 k - 1 = 4 + 2 k - 3 k^{2}$
$\Rightarrow 5 k^{2} = 5$
$\Rightarrow k = \pm 1$
$∴ A)$ Answer.

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