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

तीन दिए गए बिंदुओं \(P , Q , R\) में \(P (5,3)\) है तथा \(R\), \(x\)-अक्ष पर स्थित है। यदि \(RQ\) का समीकरण \(x-2 y=2\) है तथा \(PQ , x\)-अक्ष के समांतर है, तो \(\triangle PQR\) का केंद्रक जिस रेखा पर स्थित है, वह है

  1. A \(2x+y- 9 = 0\)
  2. B \(x - 2y+ 1 = 0\)
  3. C \(5x - 2y= 0\)
  4. D \(2x-5y = 0\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2x-5y = 0\)

Step-by-step Solution

Detailed explanation

Equation of \(RQ\) is \(x-2y=2\) .....(1) at \(Y=0\),\(X=2\) \([R (2,0)]\) as \(PQ\) is parallel to \(x,y\)-coordinates of \(Q\) is also \(3\) putting value of \(y\) in equation \((1)\), we get \(Q(8,3)\) Centroid of…
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