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

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9.Straight Line

hard

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

$a^2 -9a+ 12 =0$

(JEE MAIN-2013)

Solution

Since three lines $x-3y=p$,

$ax+2y=q$ and $ax+y=r$

from aright angled triangle

$\therefore $ product of sloper of any two lines $=-1$

Suppose $ax+2y=q$ and $x-3y=p$ are $ \bot $ to each other.

$\therefore \frac{{ – a}}{2} \times \frac{1}{3} = – 1 \Rightarrow a = 6$

Now, consider option one by one $a=6$ satisfies noly option $(a)$

$\therefore $ Required answer is ${a^2} – 9a + 18 = 0$

Standard 11

Mathematics

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