Q88 / 160 · AP EAMCET 2013 · Mathematics

$$ \begin{aligned} & \text { If the } \\ & a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \end{aligned} \text { equation } $$ represents a pair of straight lines, then the square of the distance of their point of intersection from the origin is

  • A
    $\frac{c(a+b)-a f^2-b g^2}{a b-h^2}$
  • B
    $\frac{c(a+b)+f^2+g^2}{a b-h^2}$
  • Correct answer
    C
    $\frac{c(a+b)-f^2-g^2}{a b-h^2}$
  • D
    $\frac{c(a+b)-f^2-g^2}{\left(a b-h^2\right)^2}$

Solution

We know that the point of intersection of the pair of straight line is $$ \left(\sqrt{\frac{f^2-b c}{h^2-a b}}, \sqrt{\frac{g^2-a c}{h^2-a b}}\right) $$ Required distance $$ =\left[\sqrt{\left(\sqrt{\frac{f^2-b c}{h^2-a b}}-0\right)^2+\left(\sqrt{\frac{g^2-a c}{h^2-a b}}-0\right)^2}\right]^2 $$ $\begin{aligned} & =\frac{f^2-b c}{h^2-a b}+\frac{g^2-a c}{h^2-a b} \\ & =\frac{f^2+g^2-c(a+b)}{h^2-a b} \\ & =\frac{c(a+b)-f^2-g^2}{a b-h^2}\end{aligned}$

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

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