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Find the lengths of the medians of a triangle whose vertices are $A (-1, 3), B (1, -1)$ and $C (5, 1)$.
Given:
$A (-1, 3), B (1, -1)$ and $C (5, 1)$ 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 $D=(\frac{5+1}{2}, \frac{1-1}{2})=(3 ,0)$
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 \( \mathrm{AD}=\sqrt{(3+1)^{2}+(0-3)^{2}}=\sqrt{(4)^{2}+(-3)^{2}} \)
\( =\sqrt{16+9} \)
\( =\sqrt{25} \)
\( =5 \) units
Similarly,
Coordinates of $E=(\frac{-1+5}{2}, \frac{3+1}{2})=(2 ,2)$
The length of the median \( \mathrm{BE}=\sqrt{(2-1)^{2}+(2+1)^{2}}=\sqrt{(1)^{2}+(3)^{2}} \)
\( =\sqrt{1+9} \)
\( =\sqrt{10} \) units
Coordinates of $F=(\frac{-1+1}{2}, \frac{3-1}{2})=(0, 1)$
The length of the median \( \mathrm{CF}=\sqrt{(5-0)^{2}+(1-1)^{2}}=\sqrt{(5)^{2}+(0)^{2}} \)
\( =\sqrt{25} \)
\( =5 \) units
The lengths of the medians are $5, \sqrt{10}$ and $5$ units.
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