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
- A 5
- B \(\sqrt{5}\)
- C \(\sqrt{20}\)
- D 20
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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