Question

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

• A. \(26\)
• B. \(25\)
• C. \(24\)
• D. \(34\)

Hint:

For three-dimensional coordinate geometry, use the centroid formula coordinate-wise and equate it with the given midpoint coordinate-wise.

Answer 1

The Correct Option is D

Step 1: Find the centroid of the triangle.
The centroid of a triangle with vertices \[ (x_1,y_1,z_1),\quad (x_2,y_2,z_2),\quad (x_3,y_3,z_3) \] is \[ \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right). \] Here, the vertices are \[ (4,p,-3),\quad (-1,-1,2),\quad (3,5,-8). \] So, centroid is \[ \left( \frac{4-1+3}{3}, \frac{p-1+5}{3}, \frac{-3+2-8}{3} \right). \] \[ = \left( 2, \frac{p+4}{3}, -3 \right). \]

Step 2: Find the midpoint of the given points.
The midpoint of \[ (1,4,-2) \] and \[ (q,2,-4) \] is \[ \left( \frac{1+q}{2}, \frac{4+2}{2}, \frac{-2-4}{2} \right). \] \[ = \left( \frac{1+q}{2}, 3, -3 \right). \]

Step 3: Equate the centroid and midpoint.
Since both are equal, \[ \left( 2, \frac{p+4}{3}, -3 \right) = \left( \frac{1+q}{2}, 3, -3 \right). \] Comparing \(x\)-coordinates, \[ 2=\frac{1+q}{2} \] \[ 4=1+q \] \[ q=3. \] Comparing \(y\)-coordinates, \[ \frac{p+4}{3}=3 \] \[ p+4=9 \] \[ p=5. \]

Step 4: Find \(p^2+q^2\).
\[ p^2+q^2=5^2+3^2 \] \[ =25+9 \] \[ =34 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{34} \]