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)

$a^2 - 9a + 18 = 0$

Answer 2

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 $x - 3y = pare \,\bot$ to each other.
$\therefore \quad \frac{-a}{2}\times\frac{1}{3} = -1 \Rightarrow a = 6$
Now, consider option one by one
$a = 6$ satisfies only option
$\therefore$ Required answer is $a^{2} - 9a +18 = 0$