If a triangle ABC with two vertices A(5,4,6) and B(1,–1,3) has its centroid at 103,2,113, then the third vertex C is

If G is centroid of a triangle ABC then G=A+B+C3It implies that 3G=A+B+CC=3G-A-BHence, C=10-5-1.6-4+1,11-6-3 =4,3,2

If a triangle ABC with two vertices A(5,4,6) and B(1,–1,3) has its centroid at 103,2,113, then the third vertex C is

If G is centroid of a triangle ABC then G=A+B+C3It implies that 3G=A+B+CC=3G-A-BHence, C=10-5-1.6-4+1,11-6-3 =4,3,2

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If a triangle ABC with two vertices A(5,4,6) and B(1,–1,3) has its centroid at $(\frac{10}{3}, 2, \frac{11}{3}),$ then the third vertex C is

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(4, 2, 3)

(–4, –3, 2)

(4, 3, 2)

(2, 4, 3)

answer is C.

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Detailed Solution

If $G$ is centroid of a triangle $A B C$ then $G = \frac{A + B + C}{3}$

It implies that

$\begin{array}{rcl} 3 G & = & A + B + C \\ C & = & 3 G - A - B \end{array}$

Hence,

$\begin{array}{rcl} C & = & (10 - 5 - 1.6 - 4 + 1, 11 - 6 - 3) \\ = & (4, 3, 2) \end{array}$

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If a triangle ABC with two vertices A(5,4,6) and B(1,–1,3) has its centroid at 103,2,113, then the third vertex C is

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If a triangle ABC with two vertices A(5,4,6) and B(1,–1,3) has its centroid at $(\frac{10}{3}, 2, \frac{11}{3}),$ then the third vertex C is