Question

If $a{x}^2+5xy-6y^2-10x+11y+c=0$ represents a pair of perpendicular lines then the value of $c$ is

• A. $2$
• B. $-2$
• C. $4$
• D. $-4$

Answer 1

The correct option is C $4$

$a{x}^2+5xy-6y^2-10x+11y+c=0$

It represents a pair of lines if $\Delta =0$

$\Rightarrow \left|\begin{array}{ccc}a& \frac{5}{2}& -5\\ \frac{5}{2}& -6& \frac{11}{2}\\ -5& \frac{11}{2}& c\end{array}\right|=0\Rightarrow a(-6c-\frac{121}{4})-\frac{5}{2}(\frac{5c}{2}+\frac{55}{2})-5(\frac{55}{4}-30)=0\Rightarrow -6ac-\frac{121a}{4}-\frac{25c}{4}-\frac{275}{4}-\frac{275}{4}+150=0$
$\Rightarrow \frac{121a}{4}+\frac{25c}{4}+6ac+\frac{275}{2}-150=0$ ...(1)

It represents pair of perpendicular lines if,
$a-6=0\Rightarrow a=6$

Substituting this in (1) and solving, we get
$c=4$

Hence, option C is correct answer.