TS EAMCET · Maths · Straight Lines
If \(\mathrm{L}_1\) is a line passing through the point \(\mathrm{P}(4,-3)\) and perpendicular to the line \(3 x-4 y+k=0\), then the distance of \(P\) from the line \(5 x-3 y-2=0\) measured along the line \(\mathrm{L}_1\) is
- A \(5\)
- B \(\sqrt{13}\)
- C \(\sqrt{41}\)
- D \(13\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
Slope of line perpendicular to \(3 x-4 y+k=0\) is \(\frac{-4}{3}\) \[ \therefore L_1: y+3=\frac{-4}{3}(x-4) \Rightarrow 3 y+9=-4 x+16 \] \(L_1: 4 x+3 y=7\) and \(L: 5 x-3 y=2\) (given) Point of intersection of \(L_1\) and \(L:(x, y) \equiv(1,1)\) \(\therefore \quad\) Distance…
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