Question

Find the lengths of the medians of a $△ABC$ having vertices at $A(5,1),B(1,5)$ and $C(-3,-1)$

Answer 1

$A(5,1),B(1,5)andC(-3,-1)$ are the vertices of $△ABCD$

AD,BE and CF are medians.

We have mid point formula,

$P(X,Y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Coordinate of point $D(\frac{1-3}{2},\frac{5-1}{2})=(-1,2)\therefore AD=\sqrt{(-1-5{)}^2,(2-1{)}^2}=\sqrt{36+1}=\sqrt{37}\text{Units}$

Coordinate of point $E(\frac{5-3}{2},\frac{1-1}{2})=(1,0)BE=\sqrt{(1-1{)}^2(0-5{)}^2}=\sqrt{25}=5\text{units}$

Coordinate of point $F(\frac{5+1}{2},\frac{1+5}{2})=(3,3)\therefore \sqrt{(-3-3{)}^2+(-1-3{)}^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\text{units}$