If the equation ax + 2hxy +by +2gx+2 fy +c = 0 represents two straight lines, find the square of the distance of their point of intersection from the origin. [Here you will choose numeric values of a, h, b, g, f, c as your choice like 2x? + 6xy +5 y +4x+3y+2 0 but not exactly this].

Step:-1

Given that  $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$, ---- (A),   represents two straight lines, then the square of the distance of their point of intersection from origin is $\frac{c(a+b)-f^2-g^2}{ab-h^2}$

Let $l_{1}x+m_{1}y+n_1=0 and l_{2}x+m_{2}y+n_2=0$ be the pair of straight lines represented by $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ that is, 

$\left(l_{1}x+m_{1}y+n_1\right)\left( l_{2}x+m_{2}y+n_2\right)$=$\left(a{x}^2+2hxy+b{y}^2+2gx+2fy+c\right)$

on comparing coefficient of both sides, we get

 $l_1{l}_2=a, m_1{m}_2=b, n_1{n}_2=c, m_1{n}_2+m_2{n}_1=2f, n_1{l}_2+n_2{l}_1=2g, l_1{m}_2+m_1{l}_2=2h$

Step:-2

Now, point of intersection of lines  $l_{1}x+m_{1}y+n_1=0 and l_{2}x+m_{2}y+n_2=0$ are

$\left(\frac{m_1{n}_2-m_2{n}_1}{m_1{l}_2-m_2{l}_1}, \frac{l_2{n}_1-l_1{n}_2}{m_1{l}_2-m_2{l}_1}\right)$

Note:- we can find these points by solving cross multiplication method.