Question

If the middle points of the sides of a triangle be $(-2,3),(4,-3)$ and $(4,5),$ then the centroid of triangle is:

• A. $(\frac{5}{3},2)$
• B. $(\frac{5}{6},1)$
• C. $(1,\frac{5}{6})$
• D. $(2,\frac{5}{3})$

Answer 1

Answer: (D)

$(2,\frac{5}{3})$

$(-2,3),(4,-3)$ and $(4,5),$

Given:- The vertices of $\Delta ABC$ are $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3).$

The mid point of $AB=P(-2,3),BC=Q(4,-3)$ and $AC=R(4,5).$

To find out the coordinates of $A,B$ and $C$

Let us apply mid point formula.

$\therefore P(-2,3)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},)$

$\Rightarrow x_1+x_2=-4\&y_1+y_2=6,$

$Q(4,-3)=(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2},)$

$\Rightarrow x_2+x_3=8\&y_2+y_3=-6,$ and $R(4,5)=(\frac{x_3+x_1}{2},\frac{y_3+y_1}{2},)$

$\Rightarrow x_3+x_1=8\&y_3+y_1=10.$

Adding the above equations,

${2(x}_1+x_2+x_3)=12$

$\Rightarrow x_1+x_2+x_3=6$ and $2(y_1+y_2+y_3)=10$

$\Rightarrow (y_1+y_2+y_3)=5.$

We know that the coordinates of the centroid is

$G(x,y)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})=(\frac{6}{3},\frac{5}{3})=(2,\frac{5}{3}).$