If the pairs of lines  ${\mathrm{x}}^2+2\mathrm{xy}+{\mathrm{ay}}^2=0\text{ and }{\mathrm{ax}}^2+2\mathrm{xy}+{\mathrm{y}}^2=0$ have exactly one line in common, then the joint equation of the other two lines is given by

Let y=mx be a line common to the given pairs of lines. Then, ${\mathrm{am}}^2+2\mathrm{m}+1=0$ and ${\mathrm{m}}^2+2\mathrm{m}+\mathrm{a}=0$
$\therefore \frac{{\mathrm{m}}^2}{2(1-\mathrm{a})}=\frac{\mathrm{m}}{{\mathrm{a}}^2-1}=\frac{1}{2(1-\mathrm{a})}$
Or ${\mathrm{n}}^2=1\text{ and }\mathrm{m}=-\frac{\mathrm{a}+1}{2}$
$\begin{array}{r}\mathrm{or} (\mathrm{a}+1{)}^2=4\\ \mathrm{or} \mathrm{a}=1\text{ or }-3\end{array}$
But for , the two pairs have both the lines common. So a=-3  and the slope  of the line common to both the pairs is 1. Now,
$\therefore {\mathrm{x}}^2+2\mathrm{xy}+{\mathrm{ay}}^2={\mathrm{x}}^2+2\mathrm{xy}-3{\mathrm{y}}^2=(\mathrm{x}-\mathrm{y})(\mathrm{x}+3\mathrm{y})$
And ${\mathrm{ax}}^2+2\mathrm{xy}+{\mathrm{y}}^2=-3{\mathrm{x}}^2+2\mathrm{xy}+{\mathrm{y}}^2=-(\mathrm{x}-\mathrm{y})(3\mathrm{x}+\mathrm{y})$
So, the equation of the required pair of lines is
$(\mathrm{x}+3\mathrm{y})(3\mathrm{x}+\mathrm{y})=0$
Or  $3{\mathrm{x}}^2+10\mathrm{xy}+3{\mathrm{y}}^2=0$