JEE Mains · Maths · STD 11 - 9. straight line
Given three points \(P, Q, R\) with \(P(5, 3)\) and \(R\) lies on the \(x-\) axis. If equation of \(RQ\) is \(x - 2y = 2\) and \(PQ\) is parallel to the \(x-\) axis, then the centroid of \(\Delta PQR\) lies on the line
- A \(2x+y- 9 = 0\)
- B \(x - 2y+ 1 = 0\)
- C \(5x - 2y= 0\)
- D \(2x-5y = 0\)
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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