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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Straight Lines

Let the three sides of a triangle are on the lines 4 x-7 y+10=0,...

Let the three sides of a triangle are on the lines 4x−7y+10=0,x+y=54 x-7 y+10=0, x+y=5 and 7x+4y=157 x+4 y=15. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x=0,y=0x=0, y=0 and x+y=1x+y=1 isLet 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 itsLet 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 itsLet 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

Options

  1. A5
  2. B5\sqrt{5}
  3. C20\sqrt{20}
  4. D20

Correct answer

B. 5\sqrt{5}

Step-by-step solution

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
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
<br/>7x+4y=157x+7y+10=0<br/>\begin{gathered}<br/>7 x+4 y=15 \\ 7 x+7 y+10=0<br/>\end{gathered}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
distance between P and B=5\mathrm{B}=\sqrt{5}
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