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 Δ PQR lies on the line :

Question

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 Δ PQR lies on the line : (A) 2x + y — 9 = 0 (B) x - 2y +1 = 0 (C) 5x - 2y = 0 (D) 2x - 5y = 0

Tardigrade Admin · Accepted Answer

equation of RQ≡ x-2y=2 ⇒ R (2, 0) equation of PQ = y = 3 point of intersection of PQ and RQ x - 2(3) = 2 x=8 ⇒ R(8, 3) Centroid ((2+8+5/3), (0+3+3/3)) ≡ (5, 2) as is simplified by 2x - 5y = 0

Q. Given three points $P, Q, R$ with $P (5, 3)$ and $R$ lies on the $x$ -axis. If equation of $RQ$ is $x —2 y = 2$ and $PQ$ is parallel to the $x$ -axis, then the centroid of $Δ PQR$ lies on the line :

2x + y — 9 = 0

x - 2y +1 = 0

5x - 2y = 0

2x - 5y = 0

equation of $RQ \equiv x - 2 y = 2$
$\Rightarrow R (2, 0)$
equation of $PQ = y = 3$
point of intersection of $PQ$ and $RQ$
$x - 2 (3) = 2$
$x = 8$
$\Rightarrow R (8, 3)$
Centroid $(\frac{2 + 8 + 5}{3}, \frac{0 + 3 + 3}{3})$
$\equiv (5, 2)$ as is simplified by $2 x - 5 y = 0$

Q. 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 ΔPQR lies on the line :