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 ∆PQR is a line:
(A) with slope 3/2
(B) parallel to x-axis
(C) with slope 2/3
(D) parallel to y-axis

Answer 1

Answer:

The locus of the centroid of ∆PQR is a line with slope 2/3.

Step-by-step explanation:

Let, a point P(x,y) moves on the line $2x-3y+4 = 0$.

Given that, Q(1,4) and R(3,-2) are fixed point.

Again, Let the centroid of ∆PQR is (h,k).

So, $(h,k)= (\frac{x+1+3}{3} +\frac{y+4-2}{3})$

           $h=\frac{x+4}{3}$⇒$x = 3h -4$

           $k = \frac{y+2}{3}$⇒$y = 3k - 2$

Point P(x,y) satisfying the equation of given line.

so, putting the value of x and y in the equation of line,

$2(3h-4) -3(3k-2) +4 =0$

$6h-8-9k+6+4 = 0$

$6h-9k+2 = 0$

$6x-9y+2 =0$

$y = (\frac{2}{3} )x + \frac{2}{9}$, this is the locus of the centroid of ∆PQR which is the line with slope (2/3).