Question

If the three lines $x-3y=p,ax+2y=q$ and $ax+y=r$ form a right-angled triangle then :

• A. $a^2-9a+18=0$
• B. $a^2-6a-12=0$
• C. $a^2-6a-18=0$
• D. $a^2-9a+12=0$

Answer 1

Answer: (A)

Since three lines $x-3y=p$,
$ax+2y=q$ and $ax+y=r$
form a right angled triangle
$\therefore$ product of slopes of any two lines $=-1$
Suppose $ax+2y=q$ and $x-3y=pare\perp$ to each other.
$\therefore \frac{-a}{2}\times \frac{1}{3}=-1\Rightarrow a=6$
Now, consider option one by one
$a=6$ satisfies only option
$\therefore$ Required answer is $a^2-9a+18=0$