Question

Let $A(-1,2)$ and $D(3,4)$ be the end points of the median AD of $\Delta ABC$. Find the centroid of $\Delta ABC$

• A. $(5,10)$
• B. $(5,3)$
• C. $\left(\frac{10}{3},0\right)$
• D. $\left(\frac{5}{3},\frac{10}{3}\right)$

Answer 1

The correct option is D $\left(\frac{5}{3},\frac{10}{3}\right)$

Given two point A(-1,2) and D(3,4)

AD is the median $△$ABC

Taking D as the mid point of BC

$\frac{x_2+x_3}{2}=3,\frac{y_2+y_3}{2}=4\Rightarrow x_2+x_3=6y_2+y_3=8$

Put $x_2=3\therefore x_3=3$

Put $y_2=3\therefore y_3=5$

$\therefore$ The coordinate of base (3,3) and C(3,5)

$\therefore$ Centroid of $△$ABC $=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

$=(\frac{-1+3+3}{3},\frac{2+3+5}{3})=(\frac{5}{3},\frac{10}{3})$