Question

If the equation
$12 x^{2}+7 x y-p y^{2}-18 x+q y+6=0$
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
$\begin{vmatrix}a & h & g \\h & b & f \\g & f & c\end{vmatrix}=0$ and $a +b=0$
Given pair of line is
$12 x^{2}+7 x y-p y^{2}-18 x+ q y+6=0$
$\therefore \begin{vmatrix}12 & 7 / 2 & -9 \\ 7 / 2 & -p & q / 2 \\ -9 & q / 2 & 6\end{vmatrix}=0$
and $12-p=0$
$\Rightarrow p=12$
$\therefore \begin{vmatrix}12 & 7 / 2 & -9 \\ 7 / 2 & -12 & q / 2 \\ -9 & q / 2 & 6\end{vmatrix}=0$
$\Rightarrow 12\left(-72+\frac{q^{2}}{4}\right)-\frac{7}{2}\left(-21+\frac{9 q}{2}\right) $
$-9\left(\frac{7 q}{4}-108\right)=0$
$\Rightarrow -864-3 q^{2}-\frac{147}{2}-\frac{63 q}{4}$
$-\frac{63 q}{4}+972=0$
$\Rightarrow q=1$