The vertices of a triangle are A(2, 1, 0), B(−1, 4, 3) and C(5, k, 3). If the centroid of the triangle lies on the line \(\displaystyle\frac{x-1}{1}=\frac{y-3}{2}=\frac{z-1}{1}\), then k is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 10
The centroid is the average of the vertices:
\( \displaystyle G = \left(\frac{2-1+5}{3},\ \frac{1+4+k}{3},\ \frac{0+3+3}{3}\right) = \left(2,\ \frac{5+k}{3},\ 2\right) \)
The general point of the line is \((1+\lambda,\ 3+2\lambda,\ 1+\lambda)\). Matching x = 2:
\(1+\lambda = 2\), so \(\lambda = 1\)
This gives z = 2, as needed, and the y-coordinate must be \(3+2 = 5\):
\( \displaystyle \frac{5+k}{3} = 5 \)
\(k = 10\)