Question

The points (6, 6), (0, 6) and (6, 0) are the vertices of a right triangle as shown in the figure. Find the distance between its centroid and D (point of median of AB).

• A. $\sqrt{3}\text{units}$
• B. $\sqrt{6}\text{units}$
• C. $\sqrt{2}\text{units}$
• D. $\sqrt{5}\text{units}$

Answer 1

The correct option is C

$\sqrt{2}\text{units}$

Point D = midpoint of AB
$=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$=(\frac{6+0}{2},\frac{6+0}{2})=(3,3)$

$\text{Centroid (G)}=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})=(\frac{6+0+6}{3},\frac{0+6+6}{3})=(4,4)$

$\text{Distance between two points}(x_1,y_1)\text{and}(x_2,y_2)\text{is}\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$\therefore \text{Distance between centroid and point}\text{of median}AB=\sqrt{(4-3{)}^2+(4-3{)}^2}=\sqrt{1+1}=\sqrt{2}\text{units}$