Question

lf $p{x}^2-y^2+3x+11y+q=0$ represents a pair of perpendicular lines then $(p,q)=$

• A. $(-1,-14)$
• B. $(-1,28)$
• C. $(1,-28)$
• D. $(-1,-28)$

Answer 1

Answer: (C)

$(1,-28)$

$p{x}^2-y^2+3x+11y+q=0$ represents pair of lines when
$\left|\begin{array}{ccc}p& 0& \frac{3}{2}\\ 0& -1& \frac{11}{2}\\ \frac{3}{2}& \frac{11}{2}& q\end{array}\right|=0$
$\Rightarrow p(-q-\frac{121}{4})+\frac{3}{2}\left(\frac{3}{2}\right)=0\Rightarrow -pq-\frac{121}{4}+\frac{9}{4}=0\Rightarrow pq=-28$
And the lines are perpendicular when,
$a+b=0\Rightarrow p-1=0\Rightarrow p=1$
Hence $(p,q)=(1,-28)$