A point P moves on the line 2x - 3y + 4 = 0. If Q(1, 4) and R(3, -2) are fixed points, then the locus of the centroid of ΔPQR is a line :

Question

A point P moves on the line 2x - 3y + 4 = 0. If Q(1, 4) and R(3, -2) are fixed points, then the locus of the centroid of ΔPQR is a line : (A) parallel to x-axis (B) with slope (2/3) (C) with slope (3/2) (D) parallel to y-axis

Tardigrade Admin · Accepted Answer

Let the centroid of Δ PQR is ( h , k ) P is (α, β), then (α+1+3/3)= h and (β+4-2/3)= k α=(3 h -4) β=(3 k -4) Point P (α, β) lies on line 2 x -3 y +4=0 ∴ 2(3 h -4)-3(3 k -2)+4=0 ⇒ locus is 6 x-9 y+2=0

Q. A point P moves on the line $2 x - 3 y + 4 = 0$ . If $Q (1, 4)$ and $R (3, - 2)$ are fixed points, then the locus of the centroid of $Δ$ PQR is a line :

parallel to x-axis

with slope $\frac{2}{3}$

with slope $\frac{3}{2}$

parallel to y-axis

Let the centroid of $Δ PQR$ is $(h, k) & P$ is $(α, β)$ , then
$\frac{α + 1 + 3}{3} = h$ and $\frac{β + 4 - 2}{3} = k$
$α = (3 h - 4) β = (3 k - 4)$
Point $P (α, β)$ lies on line $2 x - 3 y + 4 = 0$
$∴ 2 (3 h - 4) - 3 (3 k - 2) + 4 = 0$
$\Rightarrow$ locus is $6 x - 9 y + 2 = 0$

Q. A point P moves on the line 2x−3y+4=0. If Q(1,4) and R(3,−2) are fixed points, then the locus of the centroid of ΔPQR is a line :

parallel to x-axis

with slope 32​

with slope 23​

parallel to y-axis

Let the centroid of ΔPQR is (h,k)&P is (α,β), then 3α+1+3​=h and 3β+4−2​=k α=(3h−4)β=(3k−4) Point P(α,β) lies on line 2x−3y+4=0 ∴2(3h−4)−3(3k−2)+4=0 ⇒ locus is 6x−9y+2=0

Q. A point P moves on the line $2 x - 3 y + 4 = 0$ . If $Q (1, 4)$ and $R (3, - 2)$ are fixed points, then the locus of the centroid of $Δ$ PQR is a line :

with slope $\frac{2}{3}$

with slope $\frac{3}{2}$

Let the centroid of $Δ PQR$ is $(h, k) & P$ is $(α, β)$ , then
$\frac{α + 1 + 3}{3} = h$ and $\frac{β + 4 - 2}{3} = k$
$α = (3 h - 4) β = (3 k - 4)$
Point $P (α, β)$ lies on line $2 x - 3 y + 4 = 0$
$∴ 2 (3 h - 4) - 3 (3 k - 2) + 4 = 0$
$\Rightarrow$ locus is $6 x - 9 y + 2 = 0$