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Let each of the equations $x^{2}+2 x y+a y^{2}=0 \& a x^{2}+2 x y+y^{2}=0$ represent two straight lines passing through the origin. If they have a common line, then the other two lines are given by
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Verified Answer
The correct answer is:
$x+3 y=0,3 x+y=0$
Hint:
$\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{2}+2\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+\mathrm{a}=0 \& \mathrm{a}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{2}+2\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1=0$ have exactly one root in common (taking $\frac{\mathrm{x}}{\mathrm{y}}$ as a single
variable).
By, $\left(a_{1} b_{2}-a_{2} b_{1}\right)\left(b_{1} c_{2}-b_{2} c_{1}\right)=\left(a_{1} c_{2}-a_{2} c_{1}\right)^{2}$
We get $: \Rightarrow \mathrm{a}=1$ or $-3$
a cannot be 1
Taking $\mathrm{a}=-3$, roots of 1 st equation $: 1,-3$ and 2 nd equation : $1,-\frac{1}{3}$
So other lines: $\frac{x}{y}=-3$ and $\frac{x}{y}=-\frac{1}{3}$
$\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{2}+2\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+\mathrm{a}=0 \& \mathrm{a}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{2}+2\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1=0$ have exactly one root in common (taking $\frac{\mathrm{x}}{\mathrm{y}}$ as a single
variable).
By, $\left(a_{1} b_{2}-a_{2} b_{1}\right)\left(b_{1} c_{2}-b_{2} c_{1}\right)=\left(a_{1} c_{2}-a_{2} c_{1}\right)^{2}$
We get $: \Rightarrow \mathrm{a}=1$ or $-3$
a cannot be 1
Taking $\mathrm{a}=-3$, roots of 1 st equation $: 1,-3$ and 2 nd equation : $1,-\frac{1}{3}$
So other lines: $\frac{x}{y}=-3$ and $\frac{x}{y}=-\frac{1}{3}$
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