JEE Mains · Maths · STD 11 - 9. straight line
Let \(A(a, b), B(3,4)\) and \((-6,-8)\) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point \(P(2 a+3,7 b+5)\) from the line \(2 x+3 y-4=0\) measured parallel to the line \(x-2 y-1=0\) is
- A \(\frac{15 \sqrt{5}}{7}\)
- B \(\frac{17 \sqrt{5}}{6}\)
- C \(\frac{17 \sqrt{5}}{7}\)
- D \(\frac{\sqrt{5}}{17}\)
Answer & Solution
Correct Answer
(C) \(\frac{17 \sqrt{5}}{7}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{A}(\mathrm{a}, \mathrm{b}), \quad \mathrm{B}(3,4), \quad \mathrm{C}(-6,-8)\) \(\Rightarrow \mathrm{a}=0, \mathrm{~b}=0 \quad \Rightarrow \mathrm{P}(3,5)\) Distance from \(P\) measured along \(x-2 y-1=0\) \(\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta\) Where…
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