Question

If the centroid of a triangle with vertices \((4, p, -3), (-1, -1, 2)\) and \((3, 5, -8)\) is given by mid point of \((1, 4, -2)\) and \((q, 2, -4)\), then \(p^2+q^2 = \)

• A. 26
• B. 25
• C. 24
• D. 34

Hint:

Centroid of a triangle with vertices \((x_1,y_1,z_1), (x_2,y_2,z_2), (x_3,y_3,z_3)\) is \((\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})\).
Midpoint of two points \((x_a,y_a,z_a)\) and \((x_b,y_b,z_b)\) is \((\frac{x_a+x_b}{2}, \frac{y_a+y_b}{2}, \frac{z_a+z_b}{2})\).
Equate the coordinates of the centroid calculated in two different ways to solve for unknowns.

Answer 1

The Correct Option is D

Let the vertices of the triangle be \(A=(4, p, -3)\), \(B=(-1, -1, 2)\), and \(C=(3, 5, -8)\). The centroid \(G\) of \(\triangle ABC\) is given by: \(G = \left( \frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3}, \frac{z_A+z_B+z_C}{3} \right)\) \(G_x = \frac{4+(-1)+3}{3} = \frac{6}{3} = 2\). \(G_y = \frac{p+(-1)+5}{3} = \frac{p+4}{3}\). \(G_z = \frac{-3+2+(-8)}{3} = \frac{-9}{3} = -3\). So, the centroid is \(G = (2, \frac{p+4}{3}, -3)\). The centroid is also the midpoint of points \(P_1=(1, 4, -2)\) and \(P_2=(q, 2, -4)\). Midpoint \(M\) of \(P_1P_2\) is: \(M = \left( \frac{1+q}{2}, \frac{4+2}{2}, \frac{-2+(-4)}{2} \right) = \left( \frac{1+q}{2}, \frac{6}{2}, \frac{-6}{2} \right) = \left( \frac{1+q}{2}, 3, -3 \right)\). Since \(G=M\), we equate the coordinates: x-coordinate: \(2 = \frac{1+q}{2} \Rightarrow 4 = 1+q \Rightarrow q = 3\). y-coordinate: \(\frac{p+4}{3} = 3 \Rightarrow p+4 = 9 \Rightarrow p = 5\). z-coordinate: \(-3 = -3\) (Consistent). So, we found \(p=5\) and \(q=3\). We need to calculate \(p^2+q^2\). \(p^2+q^2 = (5)^2 + (3)^2 = 25 + 9 = 34\). This matches option (d). \[ \boxed{34} \]