Question

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

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

Answer 1

The correct option is C $a^2-9a+18=0$

Given : $L_1:x-3y=p,L_2:ax+2y=q$ and $L_3:ax+y=r$
slope of : $L_1=\frac{1}{3},L_2=-\frac{a}{2},L_3=-a$
For right angle triangle: $m_{L_1}\times m_{L_2}=-1$ or $m_{L_1}\times m_{L_3}=-1$
$[\text{as}{m}_{L_2}\times m_{L_3}=-1\text{not possible}]$
$\Rightarrow \frac{1}{3}\times \frac{-a}{2}=-1$ or $\frac{1}{3}\times -a=-1$
$\Rightarrow a-6=0$ or $a-3=0$
combining both $\Rightarrow (a-6)(a-3)=0$
$\Rightarrow a^2-9a+18=0$