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Find the lengths of the medians of a $\triangle ABC$ having vertices $A (0, -1), B (2, 1)$ and $C (0, 3)$.
Given:
$A (0, -1), B (2, 1)$ and $C (0, 3)$ are the vertices of a $\triangle ABC$.
To do:
We have to find the lengths of the medians.
Solution:
Let $D, E, F$ be the mid-points of $BC, AC$ and $AB$ respectively.
This implies, using mid-point formula,
Coordinates of $D=(\frac{2+0}{2}, \frac{1+3}{2})=(1,2)$
We know that,
The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).
Therefore,
The length of the median \( \mathrm{AD}=\sqrt{(1-0)^{2}+(2+1)^{2}} \)
\( =\sqrt{(1)^{2}+(3)^{2}} \)
\( =\sqrt{1+9} \)
\( =\sqrt{10} \) units
Similarly,
Coordinates of $E=(\frac{0+0}{2}, \frac{-1+3}{2})=(0,1)$
The length of the median \( \mathrm{BE}=\sqrt{(2-0)^{2}+(1-1)^{2}} \)
\( =\sqrt{(2)^{2}+(0)^{2}} \)
\( =\sqrt{4} \)
\( =2 \) units
Coordinates of $F=(\frac{2+0}{2}, \frac{1-1}{2})=(1, 0)$
The length of the median \( \mathrm{CF}=\sqrt{(1-0)^{2}+(0-3)^{2}} \)
\( =\sqrt{(1)^{2}+(-3)^{2}} \)
\( =\sqrt{10} \) units
The lengths of the medians are $\sqrt{10}, 2$ and $\sqrt{10}$ units.
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