Question

The two vertices of a triangle are A(6, 3) and B(−1, 7) and its centroid is G(1, 5). The third vertex C of ∆ABC is

(a) (2, 5)
(b) (2, −5)
(c) (−2, 5)
(d) (−2, −5)

Answer 1

(c) (−2, 5)
The two vertices of $∆ABC\text{are}A\left(6,3\right)\text{and}B\left(-1,7\right)$.
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
$$\begin{gathered} G\left(\frac{6-1+a}{3},\frac{3+7+b}{3}\right) \\ \text{i.e.}G\left(\frac{5+a}{3},\frac{10+b}{3}\right) \end{gathered}$$
But it is given that the centroid is G(1,5).
Therefore,
$$\begin{gathered} \frac{5+a}{3}=1\text{and}\frac{10+b}{3}=5 \\ \Rightarrow 5+a=3\text{and}10+b=15 \\ \Rightarrow a=-2\text{and}b=5 \end{gathered}$$
Hence, the third vertex of $∆ABC\text{is}C\left(-2,5\right).$