Question

Find $p$ and $q$ if equation $p{x}^2-8xy+3y^2+14x+2y+q=5$, represents pair of perpendicular lines.

Answer 1

Given, $p{x}^2-8xy+3y^2+14x+2y+q=5$...(i) represents pair of

perpendicular lines.

equation (i) is of the form

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

$\Rightarrow a=p,h=4,b=3,g=7,f=1,c=q-5.$

since the equation (i) represents pair of perpendicular lines

$a+b=0$

$\Rightarrow p+3=0$

$\Rightarrow p=-3....\left(ii\right)$

also eqn (i) represents a pair of straight lines

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

$\Rightarrow \left|\begin{array}{ccc}p& -4& 7\\ -4& 3& 1\\ 7& 1& q-5\end{array}\right|=0$

$\Rightarrow p(3q-15-1)+4(-4q+20-7)+7(-4-21)=0$

$\Rightarrow (-3)(3q-16)+4(-4q+13)-175=0$ [from (ii)P=-3]

$\Rightarrow -9q+48-16q+52-175=0$

$\Rightarrow -25q-75=0$

$\Rightarrow -25q=75$

$\therefore q=-3$

Hence $p=-3,q=-3.$