Let P Q R be a triangle with R(-1,4,2). Suppose M(2,1,2) is the mid point of PQ. The distance of the centroid of △ PQR from the point of intersection of the lines (x-2)/0=y/2=(z+3)/-1 and (x-1)/1=(y+3)/-3=(z+1)/1 is

Solution

Maths sir · 4 steps

You do not need P and Q: the centroid sits two-thirds of the way from R to the midpoint M.

1. Centroid from median. RM is a median, and the centroid G divides it in the ratio 2:1 from R. So G = (R + 2M)/3 = ((-1+4)/3, (4+2)/3, (2+4)/3) = (1, 2, 2).
2. Write both lines. The 0 under x-2 means x never changes, so line 1 is (2, 2t, -3-t). Line 2 is (1+s, -3-3s, -1+s).
3. Find intersection. Matching x: 1+s = 2 gives s = 1, so the point is (2, -6, 0). Check on line 1: 2t = -6 gives t = -3, and then z = -3+3 = 0, which matches.

   d = √((2-1)² + (-6-2)² + (0-2)²) = √(1 + 64 + 4) = √69

4. So. The distance between the centroid (1, 2, 2) and the intersection point (2, -6, 0) is √69, which is option (c).

The idea to keep. Given a vertex and the midpoint of the opposite side, the centroid is (vertex + 2 × midpoint)/3. A zero in a line's denominator means that coordinate stays fixed.

Answer from the NTA final answer key. Solution by PrepWiser.