If the straight lines joining the origin and the points of intersection of the curve $5 x^{2} + 12 x y - 6 y^{2} + 4 x - 2 y + 3 = 0$ and $x + k y - 1 = 0$ are equally inclined to the co-ordinate axis, then find the value of $| k |$

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Solution

$5 x^{2} + 12 x y - 6 y^{2} + 4 x - 2 y + 3 = 0$ ------ ( 1 )

$x + k y - 1 = 0$

$x + k y = 1$ ----- ( 2 )

Hamoginizing ( 1 ) with ( 2 )

$\Rightarrow$ $5 x^{2} + 12 x y - 6 y^{2} + 4 x (x + k y) - 2 y (x + k y) + 3 (x + k y)^{2} = 0$

$\Rightarrow$ $5 x^{2} + 12 x y - 6 y^{2} + 4 x^{2} + 4 k x y - 2 x y - 2 k y^{2} + 3 (x^{2} + k x y + k^{2} y^{2}) = 0$

$\Rightarrow$ $9 x^{2} + 10 x y - 6 y^{2} + 4 k x y - 2 k y^{2} + 3 x^{2} + 3 k x y + 3 k^{2} y^{2} = 0$

$\Rightarrow$ $12 x^{2} + (10 + 4 k + 6 k) x y + (3 k^{2} - 2 k - 6) y^{2} = 0$

$\Rightarrow$ $12 x^{2} + (10 + 10 k) x y + (3 k^{2} - 2 k - 6) y^{2} = 0$

Pair of lines are equally inclined to the axes.

$∴$ Coefficient of $x y = 0$

$\Rightarrow$ $10 k + 10 = 0$

$\Rightarrow$ $10 k = - 10$

$\Rightarrow$ $k = - 1$

$∴$ $| k | = | - 1 | = 1$

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