Question

Let each of the equations $x^2+2xy+a{y}^2=0$ & $a{x}^2+2xy+y^2=0$ represent two straight lines passing through the origin. If they have a common line, then the other two lines are given by

• A. $x-y=0,x-3y=0$
• B. $x+3y=0,3x+y=0$
• C. $3x+y=0,3x-y=0$
• D. $(3x-2y)=0,x+y=0$

Answer 1

The correct option is B $x+3y=0,3x+y=0$

Let $y=mx$ be a line common to the given pairs of lines. Then
$a{m}^2+2m+1=0\cdots \left(1\right)$
and $m^2+2m+a=0\cdots \left(2\right)$
Solving the above equations, we have
$\Rightarrow m^2=1$ and $m=-\frac{a+1}{2}$
$\Rightarrow (a+1{)}^2=4\Rightarrow a=1,-3$
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 $1$.
Now
$x^2+2xy+a{y}^2=x^2+2xy-3y^2=(x-y)(x+3y)$
and
$a{x}^2+2xy+y^2=-3x^2+2xy+y^2=-(x-y)(3x+y)$
So the equation of the required equation of the line is
$3x+y=0;x+3y=0$