Question

If one of the lines in the pair of straight lines given by $4x^2+6xy+k{y}^2=0$ bisects the angle between the coordinate axes, then $k\in$

• A. $\{-2,-10\}$
• B. $\{-2,10\}$
• C. $\{-10,2\}$
• D. $\{2,-10\}$

Answer 1

The correct option is D $\{2,-10\}$

$\left[\begin{array}{ccc}a& h& g\\ h& b& f\end{array}\right]=0$

$m_1+m_2=\frac{-2n}{b}$ and $m_1{m}_2=a/b=\frac{4}{k}$

$m_1+m_2=\frac{-6}{k}$ and $m_1{m}_2=\frac{4}{k}$

As one line bisects angle between co-ordinates

So, $m_1=1$ or $-1$

If $m_1=1$

$m_2=\frac{4}{k}\Rightarrow 1+\frac{4}{k}=\frac{-6}{k}$

$\Rightarrow 1=\frac{-10}{k}$

$\Rightarrow k=-10$

Now, if $m_1=-1$

$m_2=\frac{-4}{k}\Rightarrow -1-\frac{4}{k}=\frac{-6}{k}$

$\Rightarrow -1=\frac{-2}{k}$

$\Rightarrow k=2$

$\therefore k\in \{2,-10\}$