TS EAMCET · Maths · Straight Lines
A straight line \(L \equiv 0\) passing through the point \(A=(-5,-4)\) and having slope \(\tan \theta\) meets the lines \(x+3 y+2=0\) and \(2 x+y+4=0\) respectively at the points \(B\) and C. If \(\frac{100}{A C^2}-\frac{225}{A B^2}=4 \cos 2 \theta+\sin 2 \theta\), then the slope of the line \(L \equiv 0\) is
- A \(\frac{2}{3}\)
- B \(\frac{-2}{3}\)
- C \(\frac{-1}{2}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{-1}{2}\)
Step-by-step Solution
Detailed explanation
We have, \(A(-5,-4)\) and slope of \(L=0\) is \(\tan \theta\) \(\therefore \quad B=\left(-5+r_1 \cos \theta,-4+r_1 \sin \theta\right)\) and \(\quad C=\left(-5+r_2 \cos \theta,-4+r_2 \sin \theta\right)\) Since , \(B\) lies on \(x+3 y+2=0\)…
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