Question

Find the values of $p$ and $q$, if the following equation represents a pair of perpendicular lines :
$p{x}^2-8xy+3y^2+14x+2y+q=0$

Answer 1

We have,

$p{x}^2-8xy+3y^2+14x+2y+q=0$

For perpendicular lines,

$a+b=0$

$p+3=0$

$p=-3$

Therefore,

$-3x^2-8xy+3y^2+14x+2y+q=0$

$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$

$\left|\begin{array}{ccc}-3& -4& 7\\ -4& 3& 1\\ 7& 1& q\end{array}\right|=0$

$-3(3q-1)+4(-4q-7)+7(-4-21)=0$

$-9q+3-16q-28-28-147=0$

$25q=-200$

$q=-8$

Hence, this is the answer.