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, ${\mathrm{am}}^2+2\mathrm{m}+1=0\text{ and }{\mathrm{m}}^2+2\mathrm{m}+\mathrm{a}=0$ 
$\text{ or }\frac{{\mathrm{m}}^2}{2(1-\mathrm{a})}=\frac{\mathrm{m}}{{\mathrm{a}}^2-1}=\frac{1}{2(1-\mathrm{a})};\mathrm{a}=1\text{ (or) }-3$
But for a = 1, the two pairs have both the lines common. So a = –3 and the slope m of the line common to both the pairs is 1.
So, the equation of the required lines is $(\mathrm{x}+3\mathrm{y})(3\mathrm{x}+\mathrm{y})=0\text{ (or) }3{\mathrm{x}}^2+10\mathrm{xy}+3{\mathrm{y}}^2=0$