First slide

Equation of line in Straight lines

Question

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

Moderate

A

$a^{2} - 6 a - 12 = 0$
B

$a^{2} - 9 a + 12 = 0$
C

$a^{2} - 9 a + 18 = 0$
D

$a^{2} - 6 a - 18 = 0$

Solution

If x-3y=p and ax+2y=q are perpendicular, then $\frac{1}{3} . (- \frac{a}{2}) = - 1 i . e, a = 6....... (1)$
If x-3y=p and ax+y=r are perpendicular, then $\frac{1}{3} . (- a) = - 1 i . e, a = 3....... (2)$
ax+2y=q and ax+y=r cannot be perpendicular, as $- \frac{a}{2} (- a) = - 1 \Rightarrow a^{2} = - 2$ (not possible)
Thus, a = 6 or a = 3 or $a^{2} - 9 a + 18 = 0$

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