Q86 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

The equation of straight line passing through the point of intersection of the lines represented by $x^2+4 x y+3 y^2-$ $4 x-10 y+3=0$ and the point $(2,2)$ is

  • A
    $2 x+3 y-10=0$
  • Correct answer
    B
    $3 x+2 y-10=0$
  • C
    $2 x+y-6=0$
  • D
    $x+2 y-6=0$

Solution

$x^2+4 x y+3 y^2-4 x-10 y+3=0$ Comparing with $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ $\begin{aligned} & \text { Point of intersection }=\left(\frac{b g-f h}{h^2-a b}, \frac{a f-g h}{h^2-a b}\right) \\ & \Rightarrow\left(x_1, y_1\right)=\left(\frac{3(-2)-(-5) 2}{4-3}, \frac{-5-(-2) 2}{4-3}\right)=(4,-1) \end{aligned}$
Equation of lines passing $(4,-1)$ and $(2,2)$ is $y+1=\frac{2+1}{2-4}(x-4) \Rightarrow 3 x+2 y-10=0$

Hence, the correct answer is option: 'B'.

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