question_answer

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 \[DPQR\] lies on the line:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

A) \[2x+y-9=0\text{ }~\]  

B) \[x-2y+1=0\]

C) \[5x - 2y = 0\]

D) \[2x - 5y = 0\]

Correct Answer: D

Solution :

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 \[\Delta PQR=\left( \frac{8+5+2}{3},\frac{3+3}{3} \right)=(5,2)\]Only (2x ? 5y = 0) satisfy the given co-ordinates.