Question

The equation $12x^2+7xy+a{y}^2+13x-y+3=0$ represents a pair of perpendicular lines. Then the value of ${}^{\prime }{a}^{\prime }$ is

• A. $\frac{7}{2}$
• B. $-19$
• C. $-12$
• D. $12$

Answer 1

Answer: (C)

$-12$

Comparing the given equation with standard equation, we get $a=12$ and $b=a$, for perpendicular lines
coefficient of $x^2+$ coefficient of $y^2=0$
$\therefore 12+a=0$
$\Rightarrow a=-12$