TS EAMCET · Maths · Straight Lines
Let \(\mathrm{M}\) be the foot of the perpendicular drawn from the point \((5,-7)\) to the line \(3 x-5 y+1=0\). Then the perpendicular distance from \(M\) to the line \(2 x+5 y-3=0\) is
- A \(\frac{1}{2 \sqrt{29}}\)
- B \(\frac{9}{2 \sqrt{29}}\)
- C \(\frac{13}{2 \sqrt{29}}\)
- D \(\frac{3}{2 \sqrt{29}}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2 \sqrt{29}}\)
Step-by-step Solution
Detailed explanation
The equation of the line perpendicular to \(3x-5y+1=0\) and passing through \((5,-7)\) is \(5x+3y-4=0\). To find the foot of the perpendicular \(M\), solve the system: \(3x-5y=-1\) \(5x+3y=4\) \(9x-15y=-3\) \(25x+15y=20\) \(34x=17 \implies x=\frac{1}{2}\)…
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