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Q.
If $2\,x^2 - 10 \,xy + 2 \lambda y^2 + 5\,x - 16 \,y - 3 = 0 $ represents a pair of straight lines, then point of intersection of those lines is
TS EAMCET 2017
Solution:
On comparing the given equations with
$a \,x^{2}+2 \,h x y+b y^{2}+2 \,g x+2 \,f y+c=0$, 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
$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}=0 $
$\Rightarrow \begin{vmatrix} 2 & -5 & 5 / 2 \\ -5 & 2 \lambda & -8 \\ 5 / 2 & -8 & -3 \end{vmatrix}=0$
$\Rightarrow 2(-6 \lambda-64)+5(15+20)+\frac{5}{2}(40-5 \lambda)=0$
$\Rightarrow -12 \lambda-128+175+100-\frac{25}{2} \lambda=0$
$\Rightarrow -\frac{49}{2} \lambda=-147$
$\Rightarrow \lambda=\frac{147 \times 2}{49}=6$
Now, the point of intersection of given lines is given by
$\left(\frac{b g-f h}{h^{2}-a b}, \frac{a f-g h}{h^{2}-a b}\right) =\left(\frac{30-40}{25-24}, \frac{-16+\frac{25}{2}}{25-24}\right) $
$=\left(-10, \frac{-7}{2}\right) $