Prove that the straight lines joining origin to the points of intersection of the straight line $3 x - y - 2 = 0$ and the curve $7 x^{2} - 4 x y + 8 y^{2} + 2 x - 4 y - 8 = 0$ are at right angles of each other.

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Solution

$3 x - y - 2 = 0$ or $\frac{3 x - y}{2} = 1$ .
Making the equation of the curve homogenous, we get
$7 x^{2} - 4 x y + 8 y^{2} + (2 (x - 2 y) \frac{(3 x - y)}{2} - 8 \cdot {(\frac{3 x - y}{2})}^{2} = 0$
or $(7 x^{2} - 4 x y + 8 y^{2}) + (3 x^{-} 7 x y + 2 y^{2}) - 2 (9 x^{2} - 6 x y + y^{2}) = 0$
or $- 8 x^{2} + x y + 8 y^{2} = 0$ is the required equation to the lines.
Since $a + b = - 8 + 8 = 0$ , and hence the lines are perpendicular.

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