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9.Straight Line
hard
જો ત્રણ રેખા $x - 3y = p, ax + 2y = q$ અને $ax + y = r$ કાટકોણ ત્રિકોણની બાજુઓ હોય તો
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
