Find the value of $k$ , if the lines joining to the points of inter section of the curve $2 x^{2} - 2 x y + 3 y^{2} + 2 x - y - 1 = 0$ and the line $x + 2 y = k$ are mutually perpendicular.

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Solution

Given that,
the curve
$2 x^{2} - 2 x y + 3 y^{2} + 2 x - y - 1 = 0$ (1)
the line
$x + 2 y = k$
$\frac{x + 2 y}{k} = 1$ (2)
By equation (1) to
$(2 x^{2} - 2 x y + 3 y^{2}) + (2 x - y) 1 - 1 (1)^{2}$
From equation(2) to,
$(2 x^{2} - 2 x y + 3 y^{2}) + (2 x - y) \frac{(x + 2 y)}{k} - 1 {(\frac{x + 2 y}{k})}^{2} = 0$
$k^{2} (2 x^{2} - 2 x y + 3 y^{2}) + k (2 x - y) (x + 2 y) - (x + 2 y)^{2} = 0$
$2 k^{2} x^{2} - 2 k^{2} x y + 3 k^{2} y^{2} + 2 k x^{2} + 3 k x y - 2 k y^{2} - x^{2} - 4 x y - 4 y^{2} = 0$
$\Rightarrow x^{2} (2 k^{2} + 2 k - 1) + x y (- 2 k^{2} + 3 k - 4) + y^{2} (3 k^{2} - 2 k - 4) = 0$
These lines are given mutually perpendicular
$∴ x^{2} c o e f f . + y^{2} c o e f f = 0$
$\Rightarrow 2 k^{2} + 2 k - 1 + 3 k^{2} - 2 k - 4 = 0$
$\Rightarrow 5 k^{2} - 5 = 0$
$k^{2} - 1 = 0$ , $k = \pm 1$

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