Question

If $\lambda x^2-5xy+6y^2+x-3y=0$ represents a pair of straight lines, then their point of intersection is:

• A. $\left(1,3\right)$
• B. $\left(-1,-3\right)$
• C. $\left(3,1\right)$
• D. $\left(-3,-1\right)$

Answer 1

The correct option is D

$\left(-3,-1\right)$

Explanation for the correct option.

Step 1: Find the value of $\lambda$.

The general equation for pair of lines is $a{x}^2+b{y}^2+2hxy+2gx+2fy+c=0$.

The given equation is $\lambda x^2-5xy+6y^2+x-3y=0$

By comparing we get,

$a=\lambda ,b=6,h=-5/2,g=1/2,f=-3/2,andc=0$

The condition for straight lines is

$\begin{array}{rcl}abc+2fgh-a{f}^2-b{g}^2-c{h}^2& =& 0\\ & \Rightarrow & \left(\lambda \times 6\times 0\right)+2\left(\frac{-3}{2}\times \frac{1}{2}\times \frac{-5}{2}\right)-\frac{9}{4}\lambda -6\left(\frac{1}{4}\right)-0\left(\frac{25}{4}\right)=0\\ \Rightarrow \frac{15}{4}-\frac{9}{4}\lambda -\frac{6}{4}& =& 0\\ \Rightarrow \frac{9}{4}& =& \frac{9}{4}\lambda \\ \Rightarrow \lambda & =& 1\end{array}$

Step 2: Find the point of intersection.

The given equation will now be

$\begin{array}{rcl}{x}^2-5xy+6y^2+x-3y& =& 0\\ \Rightarrow x^2-3xy-2xy+6y^2+x-3y& =& 0\\ & \Rightarrow & x\left(x-3y\right)-2y\left(x-3y\right)+1\left(x-3y\right)=0\\ \Rightarrow \left(x-2y+1\right)\left(x-3y\right)& =& 0\end{array}$

From here we get

$\begin{array}{rcl}x-3y& =& 0\\ \Rightarrow x& =& 3y\end{array}$

and

$\begin{array}{rcl}x-2y+1& =& 0\\ & \Rightarrow & 3y-2y=-1\\ \Rightarrow y& =& -1\end{array}$

so, $x=-3$

Therefore, the point of intersection is $\left(-3,-1\right)$.

Hence, option D is the correct answer.