AP EAMCET · Maths · Straight Lines
If the angle between the pair of lines \(x^2+2 \sqrt{2} x y+k y^2\) \(=0, \mathrm{k}>0\) is \(45^{\circ}\), then the area (in square units) of the triangle formed by the pair of bisectors of the angles between these lines and the line \(x+2 y+1=0\) is
- A \(\frac{1}{3}\)
- B 1
- C \(\frac{2}{3}\)
- D 2
Answer & Solution
Correct Answer
(A) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
Given: \(x^2+2 \sqrt{2} x y+k y^2=0\) ...(i) \(x+2 y+1=0\) ...(ii) The angle between the lines (i) is \(45^{\circ}\). \(\therefore \cos 45^{\circ}=\left|\frac{1+k}{\sqrt{(1-k)+8}}\right| \Rightarrow \frac{1}{\sqrt{2}}=\left|\frac{1+k}{\sqrt{9-k}}\right|\)…
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