Question

The coordinates of the midpoints of the sides of a triangle are $(4,2)$, $(3,3)$ and$(2,2)$. What will be the coordinates of the centroid of the triangle?

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

Answer 1

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

Coordinates of the mid-point of A and B are

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$⟹\frac{x_1+x_2}{2}=4,\frac{y_1+y_2}{2}=2$

$\therefore x_1+x_2=8,y_1+y_2=4$ (1)

Coordinates of mid-point of B and C are

$\frac{x_2+x_3}{2}=3\to x_2+x_3=6,$ (2)

$\frac{y_2+y_3}{2}=3\to y_2+y_3=6$

Coordinates of mid-point of C and A are

$\frac{x_1+x_3}{2}=2\to x_1+x_3=4$, (3)

$\frac{y_1+y_3}{2}=2\to y_1+y_3=4$

Add the equation (1),(2) and (3)

$2(x_1+x_2+x_3)=182(y_1+y_2+y_3)=14\Rightarrow x_1+x_2+x_3=9;x_1+x_2+x_3=7$

Given that $x_1+x_2=8and{y}_1+y_2=4$

$8+x_3=9⟹x_3=14+y_3=7⟹y_3=3x_1+x_3=4⟹x_1=3y_1+y_3=4⟹y_1=1\therefore x_2=5and{y}_2=3$

$\therefore$ The vertices of$△$le A(3,1) ;B(5,3) ;C(1,3)

Centroid of triangle $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})=(\frac{9}{3},\frac{7}{3})=(3,\frac{7}{3})$