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JEE Mains · Maths · STD 11 - 9. straight line

Let the three sides of a triangle are on the lines \(4 x-7 y+10=0, x+y=5\) and \(7 x+4 y=15\). Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines \(x=0, y=0\) and \(x+y=1\) is

  1. A 5
  2. B \(\sqrt{5}\)
  3. C \(\sqrt{20}\)
  4. D 20
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Answer & Solution

Correct Answer

(B) \(\sqrt{5}\)

Step-by-step Solution

Detailed explanation

\(7 x+4 y=15\) \(7 x+7 y+10=0\) distance between P and \(\mathrm{B}=\sqrt{5}\)
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