If 2 x^2-10 x y+2 lambda y^2+5 x-16 y-3=0 rep | vedclass

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(C) Comparing the given equation 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 equat

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$\left(-10, \frac{-7}{2}\right)$

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(C) Comparing the given equation 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 equation represents a pair of straight lines,the determinant condition is:$\left|\begin{array}{ccc} a & h & g \ h & b & f \ g & f & c \end{array}ight|=0$ $\Rightarrow \left|\begin{array}{ccc} 2 & -5 & 5/2 \ -5 & 2 \lambda & -8 \ 5/2 & -8 & -3 \end{array}ight|=0$.Expanding the determinant: $2(-6 \lambda-64)+5(15+20)+\frac{5}{2}(40-5 \lambda)=0$.$-12 \lambda-128+175+100-12.5 \lambda=0$ $\Rightarrow -24.5 \lambda = -147$ $\Rightarrow \lambda=6$.Now,$b=2 \lambda = 12$. The point of intersection $(x, y)$ is given by $\left(\frac{b g-f h}{h^2-a b}, \frac{a f-g h}{h^2-a b}ight)$.Denominator $h^2-a b = (-5)^2 - (2)(12) = 25-24=1$.$x = \frac{(12)(5/2) - (-8)(-5)}{1} = 30-40 = -10$.$y = \frac{(2)(-8) - (5/2)(-5)}{1} = -16 + 12.5 = -3.5 = \frac{-7}{2}$.Thus,the point of intersection is $\left(-10, \frac{-7}{2}ight)$.

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

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