Question

If a vertex of a triangle is $(1,1)$ and the mid points of two sides through this vertex are $(-1,2)$ and $(3,2)$, then the centroid of the triangle is

• A. $\left(\frac{-1}{3},\frac{7}{3}\right)$
• B. $(-1,\frac{7}{3})$
• C. $\left(\frac{1}{3},\frac{7}{3}\right)$
• D. $(1,\frac{7}{3})$

Answer 1

The correct option is D $(1,\frac{7}{3})$

Let the vertices of the triangle be $A(1,1)$ $B(x_1,y_1)$ and $C(x_2,y_2)$.

Hence, the midpoint of $AB$ is given by $(-1,2)=(\frac{x_1+1}{2},\frac{y_1+1}{2})$.

Therefore, $x_1=-3$ and $y_1=3$.

Hence, the vertex $B$ is given by $(-3,3)$.

Similarly the midpoint of $AC$ is given by $(3,2)=(\frac{x_2+1}{2},\frac{y_2+1}{2})$

Hence, $x_2=5$ and $y_2=3$.
Thus, the vertices are $A(1,1)$, $B(-3,3)$ and $C(5,3)$.

Hence the centroid is given by,
$G=(\frac{1-3+5}{3},\frac{1+3+3}{3})=(1,\frac{7}{3})$