Here given vertices are A (0,-1), B (2, 1) and C (0, 3) and let midpoints of BC, CA and AB be D,E and F respectively.
By midpoint formula.
x = \(\frac{x_1+x_2}2\), y = \(\frac{y_1+y_2}2\)
For midpoint D of side BC,
x = \(\frac{-3+1}2\), y = \(\frac{-1+5}2\)
x = \(\frac{-2}2\), y = \(\frac{4}2\)
∴midpoint of side BC is D(-1, 2)
For midpoint E of side AB,
x = \(\frac{-3+5}2\), y = \(\frac{-1+1}2\)
x = \(\frac{2}2\), y = \(\frac{0}2\)
∴midpoint of side AB is E(1, 0)
For midpoint F of side CA,
x = \(\frac{1+5}2\), y = \(\frac{1+5}2\)
x = \(\frac{6}2\), y = \(\frac{6}2\)
∴midpoint of side CA is F(3, 3)
By distance formula,
XY = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
For median AD,
AD = \(\sqrt{(-1-5)^2+(2-1)^2}\)
= \(\sqrt{36 + 1}\)
= \(\sqrt{37}\) units
For median BE,
BE = \(\sqrt{(1-1)^2+(0-5)^2}\)
= \(\sqrt{25}\)
= 5 units.
For median CF,
CF = \(\sqrt{(-3-3)^2+(-1-3)^2}\)
= \(\sqrt{36 + 16}\)
= \(2\sqrt{13}\) units
