Write down the equation of the pair of straight lines joining the origin to the points of intersection of the line  $6\mathrm{x}-\mathrm{y}+8=0$ with the pair of straight lines $3{\mathrm{x}}^2+4\mathrm{xy}-4{\mathrm{y}}^2-11\mathrm{x}+2\mathrm{y}+6=0$ show that the lines. So obtained make equal angles with the coordinate axes.

Given equation is
$3{\mathrm{x}}^2+4\mathrm{xy}-4{\mathrm{y}}^2-11\mathrm{x}+2\mathrm{y}+6=0 ...\left(1\right)$
$\text{ and line is }6\mathrm{x}-\mathrm{y}+8=0\Rightarrow 6\mathrm{x}-\mathrm{y}=-8$
$\Rightarrow \frac{6\mathrm{x}-\mathrm{y}}{-8}=1\Rightarrow \frac{\mathrm{y}-6\mathrm{x}}{8}=1 ...\left(2\right)$
Let (1) & (2) intersect at A, B.
By homogenenizing (1) with line (2), we get the combined equation of $\overline{\mathrm{OA}}\text{ and }\overline{\mathrm{OB}}$
$\begin{array}{l}\therefore 3{\mathrm{x}}^2+4\mathrm{xy}-4{\mathrm{y}}^2-11\mathrm{x}\left(1\right)+2\mathrm{y}\left(1\right)+6(1{)}^2=0\\ \Rightarrow 3{\mathrm{x}}^2+4\mathrm{xy}-4{\mathrm{y}}^2-11\mathrm{x}\frac{(\mathrm{y}-6\mathrm{x})}{8}+2\mathrm{y}\frac{(\mathrm{y}-6\mathrm{x})}{8}+6\frac{(\mathrm{y}-6\mathrm{x}{)}^2}{64}=0\\ \\ \Rightarrow 64\times 3{\mathrm{x}}^2+64\times 4\mathrm{xy}-64\times 4{\mathrm{y}}^2-11\times 8\mathrm{x}(\mathrm{y}-6\mathrm{x})+2\times 8\mathrm{y}(\mathrm{y}-6\mathrm{x})+6\left({\mathrm{y}}^2-12\mathrm{xy}+36{\mathrm{x}}^2\right)=0\\ \\ \Rightarrow 192{\mathrm{x}}^2+256\mathrm{xy}-256{\mathrm{y}}^2-88\mathrm{xy}+528{\mathrm{x}}^2+16{\mathrm{y}}^2-96\mathrm{xy}+6{\mathrm{y}}^2-72\mathrm{xy}+216{\mathrm{x}}^2=0\\ \\ \Rightarrow 936{\mathrm{x}}^2+256\mathrm{xy}-256\mathrm{xy}-234{\mathrm{y}}^2=0\\ \Rightarrow 4{\mathrm{x}}^2-{\mathrm{y}}^2=0\end{array}$
The lines makes equal angles with the coordinate axes means
$\begin{array}{l}\mathrm{h}\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)=\mathrm{xy}(\mathrm{a}-\mathrm{b})\\ 0\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)=\mathrm{xy}(4+1)\\ 5\mathrm{xy}=0\Rightarrow \mathrm{xy}=0\\ \therefore \mathrm{x}=0\mathrm{\&}\mathrm{y}=0\end{array}$