Find the equation to the pair of lines joining the origin to the points of intersection of the curve $7{\mathrm{x}}^2-4\mathrm{xy}+8{\mathrm{y}}^2+2\mathrm{x}-4\mathrm{y}-8=0$ with the straight the 3x – y = 2 and the angle between them.

The given curve is
$7{\mathrm{x}}^2-4\mathrm{xy}+8{\mathrm{y}}^2+2\mathrm{x}-4\mathrm{y}-8=0$
and the given line is 3x – y = 2
$\Rightarrow \frac{3\mathrm{x}-\mathrm{y}}{2}=1$
The given curve (1) and the line (2) intersects $\text{ at }\mathrm{A},\mathrm{B}\text{ join }\overline{\mathrm{OA}}\text{ and }\overline{\mathrm{OB}}\text{. }$
By homogenizing the equation of the curve (1) with the line (2) we get
$\begin{array}{l}7{\mathrm{x}}^2-4\mathrm{xy}+8{\mathrm{y}}^2+2\mathrm{x}\left(1\right)-4\mathrm{y}\left(1\right)-8(1{)}^2=0\\ \Rightarrow 7{\mathrm{x}}^2-4\mathrm{xy}+8{\mathrm{y}}^2+2\mathrm{x}\left(\frac{3\mathrm{x}-\mathrm{y}}{2}\right)-4\mathrm{y}\left(\frac{3\mathrm{x}-\mathrm{y}}{2}\right)-8{\left(\frac{3\mathrm{x}-\mathrm{y}}{2}\right)}^2=0\\ \\ \Rightarrow 7{\mathrm{x}}^2-4\mathrm{xy}+8{\mathrm{y}}^2+3{\mathrm{x}}^2-\mathrm{xy}-6\mathrm{xy}+2{\mathrm{y}}^2-8\left(\frac{9{\mathrm{x}}^2-{\mathrm{y}}^2-6\mathrm{xy}}{4}\right)=0\\ \\ \Rightarrow 7{\mathrm{x}}^2-4\mathrm{xy}+8{\mathrm{y}}^2+3{\mathrm{x}}^2-\mathrm{xy}-6\mathrm{xy}+2{\mathrm{y}}^2-18{\mathrm{x}}^2-2{\mathrm{y}}^2+12\mathrm{xy}=0\\ \\ \Rightarrow -8{\mathrm{x}}^2+\mathrm{xy}+8{\mathrm{y}}^2=0\dots ..\left(3\right)\\ \Rightarrow 8{\mathrm{x}}^2-\mathrm{xy}-8{\mathrm{y}}^2=0\dots \end{array}$
$\text{ (3) represents the pair of lines }\overline{\mathrm{OA}},\overline{\mathrm{OB}}$ comparing equation (3) with
${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2=0, \mathrm{we} \mathrm{get}$
a = 8, 2h = –1, b = –8
now a + b = 8 – 8 = 0

lines are perpendicular
$\text{ Angle between the lines is }\frac{\mathrm{\pi }}{2}$