Question

If $2x^2-10xy+2\lambda y^2+5x-16y-3=0$ represents a pair of straight lines, then point of intersection of those lines is

• A. (2, -3)
• B. (5, -16)
• C. $\left(-10,\frac{-7}{2}\right)$
• D. $\left(-10,\frac{-3}{2}\right)$

Answer 1

Answer: (C)

On comparing the given equations with
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$, we get
$a=2,b=2\lambda ,h=-5,g=\frac{5}{2}$
$f=-8,c=-3$
Since, the given equations represents a pair of straight lines, therefore
$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}2& -5& 5/2\\ -5& 2\lambda & -8\\ 5/2& -8& -3\end{array}\right|=0$
$\Rightarrow 2(-6\lambda -64)+5(15+20)+\frac{5}{2}(40-5\lambda )=0$
$\Rightarrow -12\lambda -128+175+100-\frac{25}{2}\lambda =0$
$\Rightarrow -\frac{49}{2}\lambda =-147$
$\Rightarrow \lambda =\frac{147\times 2}{49}=6$
Now, the point of intersection of given lines is given by
$\left(\frac{bg-fh}{h^2-ab},\frac{af-gh}{h^2-ab}\right)=\left(\frac{30-40}{25-24},\frac{-16+\frac{25}{2}}{25-24}\right)$
$=\left(-10,\frac{-7}{2}\right)$