AP EAMCET · Maths · Straight Lines
The area of the triangle formed by the lines represented by \(3 x+y+15=0\) and \(3 x^2+12 x y-13 y^2=0\) is
- A \(\frac{15 \sqrt{3}}{2}\)
- B \(15 \sqrt{3}\)
- C \(\frac{15 \sqrt{3}}{4}\)
- D \(\frac{15}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(A) \(\frac{15 \sqrt{3}}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & 3 x+y+15=0 ...(i)\\ & 3 x^2+12 x y-13 y^2=0...(ii) \end{aligned}\) Point of intersection of (2) is \((0,0)\) From (i) and (ii),…
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