Question

lf the equation $3x^2+6xy+m{y}^2=0$ represents a pair of straight lines inclined at an angle $\pi$, then $m$ is equal to

• A. $3$
• B. $6$
• C. $9$
• D. any real number

Answer 1

Answer: (A)

$3$

Since there is no constant term, or linear term of x and y, hence we can conclude that both the lines pass through the origin.

Let the lines be

$y=m_{1}x$
$y=m_{2}x$

Since the angle between them is $\pi$.

Hence

$tan\left(\pi \right)=|\frac{m_2-m_1}{1+m_1{m}_2}|$

$tan\left(\pi \right)=0$

Therefore

$|\frac{m_1-m_2}{1+m_1{m}_2}|=0$

Hence

$m_1=m_2=m^{\prime }$

Thus the above pair of lines can be represented as

$(y-m^{\prime }x{)}^2=0$

$y^2-2m^{\prime }xy+m^{\prime 2}{x}^2=0$

Comparing with

$3x^2+6xy+m{y}^2=0$

$\to \frac{3}{m}+\frac{6}{m}xy+y^2=0$

Hence

$m^{\prime 2}=\frac{3}{m}$ and $\frac{6}{m}=-2m^{\prime }$

Therefore

$2m^{\prime 2}=-2m^{\prime }$

$2m^{\prime }(m^{\prime }+1)=0$
Since $m^{\prime }\ne 0$ hence $m^{\prime }=-1$

Therefore

$\frac{6}{m}=2$

$m=3$

Hence the above equation is

$3x^2+6xy+3y^2=0$