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JEE Mains · Maths · STD 11 - 9. straight line
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 & Solution
Correct Answer
(A) \(a^2 -9a + 18 =0\)
Step-by-step Solution
Detailed explanation
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.…
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