Question

The points $(11,9),(2,1)$ and $(2,-1)$ are the midpoints of the sides of the triangle. Then the centroid is

• A. $(-5,-3)$
• B. $(5,-3)$
• C. $(3,5)$
• D. $(5,3)$

Answer 1

Answer: (D)

Let the vertices of a triangle be $A\left(x_1,y_1\right)$
$B\left(x_2,y_2\right)$ and $C\left(x_3,y_3\right).$ Since, $(11,9),(2,-1)$
and $(2,1)$ are the mid-points of $AB,BC$ and $CA$.
$\frac{x_2+x_3}{2}=2,\frac{y_2+y_3}{2}=-1$
$\frac{x_3+x_1}{2}=2,\frac{y_3+y_1}{2}=1$
and $\frac{x_1+x_2}{2}=11,\frac{y_1+y_2}{2}=9$

$\therefore x_2+x_3=4,x_3+x_1=4,x_1+x_2=22$
$\therefore 2\left(x_1+x_2+x_3\right)=30$
$\Rightarrow x_1+x_2+x_3=15$
and $y_1+y_3=+2,y_2+y_3=-2y_1+y_2=18$
$\therefore 2\left(y_1+y_2+y_3\right)=18$
$\Rightarrow y_1+y_2+y_3=9$
$\therefore$ The centroid is
$=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$
$=\left(\frac{15}{3},\frac{9}{3}\right)=(5,3)$