TS EAMCET · Maths · Straight Lines
Let \(P\) be the point of intersection of the lines \(L_1 \equiv x-y-7=0\) and \(L_2 \equiv x+y-5=0\). \(A\left(x_1, y_1\right)\) and \(B\left(x_2, y_2\right)\) are points on the lines \(L_1=0\) and \(L_2=0\) respectively such that \(P A=3 \sqrt{2}, P B=\sqrt{2}, x_1, y_1 \geq 0, x_2, y_2 \geq 0\), then the angle made by the line segment \(A B\) at the origin is
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{2}\)
- C \(\cos ^{-1}\left(\frac{3}{4}\right)\)
- D \(\cos ^{-1}\left(\frac{9}{\sqrt{85}}\right)\)
Answer & Solution
Correct Answer
(D) \(\cos ^{-1}\left(\frac{9}{\sqrt{85}}\right)\)
Step-by-step Solution
Detailed explanation
The point of intersection of lines \(L_1=x-y-7=0\) and \(L_2 \equiv x+y-5=0\) is \((6,-1)\) Now, for point \(A\left(x_1, y_1\right)\) such that \(P A=3 \sqrt{2}\), take the equation of line \(L_1\) in symmetric form…
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