Question

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

• A. $9a^2+12a+4=0$
• B. $9a^2+6a+1=0$
• C. $a^2-6a-18=0$
• D. $a^2-9a+12=0$

Answer 1

Answer: (A), (B)

The correct options are
A $9a^2+12a+4=0$
B $9a^2+6a+1=0$

Form a ${90}^0-\Delta$

$x-3y=p\to \left(i\right),ax+2y=q\to \left(ii\right),ax+y=r\to \left(iii\right)$

Either $\left(i\right)$ & $\left(ii\right)$ or, $\left(i\right)$ & $\left(iii\right)$

Slope are $\perp$

$\Rightarrow \frac{1}{3}=\frac{-a}{2}$ , $\frac{1}{3}=-a$

$\Rightarrow (2+3a)=0$ , $(3a+1)=0$

$\Rightarrow {9a}^2+4+12a=0$ , ${9a}^2+1+6a=0$