Question

Find the lengths of the medians of a triangle whose vertices are $(3,4),(-1,2)$ and $(2,3)$

Answer 1

Let the given points of a triangle are $A(3,4),B(-1,2),C(2,3)$.

Let $D,E,F$ be a mid-points of sides $BC,CA,AB$.

The co-ordinates of $D:$

$=D[\frac{-1+2}{2},\frac{2+3}{2}]=D[\frac{1}{2},\frac{5}{2}]$

The co-ordinates of $E:$

$=E[\frac{2+3}{2},\frac{3+4}{2}]=E[\frac{5}{2},\frac{7}{2}]$

The co-ordinates of $F:$

$=F[\frac{3+(-1)}{2},\frac{4+2}{2}]=F[1,3]$

Now, $AD=\sqrt{{[\frac{1}{2}-3]}^2+{[\frac{5}{2}-4]}^2}$

$=\sqrt{{\left[\frac{-5}{2}\right]}^2+{\left[\frac{-3}{2}\right]}^2}$

$=\sqrt{\frac{25}{4}+\frac{9}{4}}$

$=\sqrt{\frac{34}{4}}units$

And $BE=\sqrt{{[\frac{5}{2}+1]}^2+{[\frac{7}{2}-2]}^2}$

$=\sqrt{{\left[\frac{7}{2}\right]}^2+{\left[\frac{3}{2}\right]}^2}$

$=\sqrt{\frac{49}{4}+\frac{9}{4}}$

$=\sqrt{\frac{58}{4}}units$

$CF=\sqrt{[1-2{]}^2+[3-3{]}^2}=1units$