Question

If the equation
$12x^2+7xy-p{y}^2-18x+qy+6=0$
represents a pair of perpendicular straight lines, then

• A. p = 12, q = -1
• B. p = -12, q = 1
• C. p = 12, q = 1
• D. p = 1, q = 1

Answer 1

Answer: (C)

The second degree equation will represent a pair of perpendicular straight lines, if
$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$ and $a+b=0$
Given pair of line is
$12x^2+7xy-p{y}^2-18x+qy+6=0$
$\therefore \left|\begin{array}{ccc}12& 7/2& -9\\ 7/2& -p& q/2\\ -9& q/2& 6\end{array}\right|=0$
and $12-p=0$
$\Rightarrow p=12$
$\therefore \left|\begin{array}{ccc}12& 7/2& -9\\ 7/2& -12& q/2\\ -9& q/2& 6\end{array}\right|=0$
$\Rightarrow 12\left(-72+\frac{q^2}{4}\right)-\frac{7}{2}\left(-21+\frac{9q}{2}\right)$
$-9\left(\frac{7q}{4}-108\right)=0$
$\Rightarrow -864-3q^2-\frac{147}{2}-\frac{63q}{4}$
$-\frac{63q}{4}+972=0$
$\Rightarrow q=1$