TS EAMCET · Maths · Pair of Lines
If the pair of straight lines given by \(A x^2+2 H x y+B y^2=0\left(H^2>A B\right)\) forms an equilateral triangle with line \(a x+b y+c=0\), then \((A+3 B)(3 A+B)\) is equal to :
- A \(H^2\)
- B \(-H^2\)
- C \(2 H^2\)
- D \(4 H^2\)
Answer & Solution
Correct Answer
(D) \(4 H^2\)
Step-by-step Solution
Detailed explanation
Given that, \(A x^2+2 H x y+B y^2=0\)...(i) and \(a x+b y+c=0\) ...(ii) Since, triangle is equilateral, then angle between the two lines is \(60^{\circ}\). Angle between pair of lines is given by \(\cos 60^{\circ}=\frac{A+B}{\sqrt{(A-B)^2+4 H^2}}\)…
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