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

माना एक त्रिभुज की दो भुजाओं के समीकरण \(3 x -2 y +6=0\) तथा \(4 x +5 y -20=0\) हैं। यदि इस त्रिभुज का लम्बकेंद्र \((1,1)\) पर है, तो इसकी तीसरी भुजा का समीकरण है

  1. A \(122y\, - \,26x\, - 1675\, = \,0\)
  2. B \(26x\, + \,61y\, + \,1675\, = \,0\)
  3. C \(122y\, + \,26x\, + 1675\, = \,0\)
  4. D \(26x\, - \,122y\, - \,1675\, = \,0\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(26x\, - \,122y\, - \,1675\, = \,0\)

Step-by-step Solution

Detailed explanation

Equation of \(AB\) is \(3x-2y+6=0\) Equation of \(AC\) is \(4x+5y-20=0\). Equation of \(BE\) is \(2x+3y-5=0\) Equation of \(CF\) is \(5x-4y-1=0\) \( \Rightarrow \) Equation of \(BC\) is \(26x-122y=1675\)
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