Let P (2,-3), Q (-2, 1) be the vertices of the triangle PQR. If the centroid of ΔPQR lies on the line 2x +3y = 1, then the locus of R is a. 2x + 3y = 9 b. 2x - 3y = 9 c. 3x + 2y = 5 d. 3x - 2y = 5

Correct answer is a. 2x + 3 y = 9.

Step-by-step explanation:

Let the coordinates of centroid be (h,k)

{h/3 , (-2+k)/3}

h = (2 – 2 + a)/3 = a/3 ---eqn 1

k= ( - 3+ 1 + b )/3 = (-2 + b)/3 eqn 2

Where (x,y) are any point on the line 2x+3y=1

3h = a and 3k + 2 = b

From 2x +3y = 1,

Then, 2h +3k = 1, 3k = 1 - 2h eqn 3

b = 1 - 2h + 2 = 3 - 2h

also b = 3 - 2a/3

b = (9 -2a)/3

3b = 9 - 2a

3b + 2a = 9

Now (x,y) satisfy the point on the line 2x+3y=9

So the locus is 2x + 3 y = 9.

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.