123
Find the length of the median AD of $\triangle ABC$ having vertices A$(0, -1)$, B$(2, 1)$ and C$(0, 3)$.
Show SolutionHide Solution↓
Coordinate of D$(1,2)$.
AD$= \sqrt{(1-0)^2 + (2 - (-1))^2}$
$= \sqrt{1^2 + 3^2} = \sqrt{1+9} = \sqrt{10}$
AD$= \sqrt{(1-0)^2 + (2 - (-1))^2}$
$= \sqrt{1^2 + 3^2} = \sqrt{1+9} = \sqrt{10}$