Find the lengths of the medians of a triangle
Question:

Find the lengths of the medians of a triangle whose vertices are A (−1,3), B(1,−1) and C(5, 1).

Solution:

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (−1, 3); B (1,−1) and C (5, 1).

So we should find the mid-points of the sides of the triangle.

In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Therefore mid-point P of side AB can be written as,

$P(x, y)=\left(\frac{-1+1}{2}, \frac{3-1}{2}\right)$

Now equate the individual terms to get,

$x=0$

 

$y=1$

So co-ordinates of P is (0, 1)

Similarly mid-point Q of side BC can be written as,

$Q(x, y)=\left(\frac{5+1}{2}, \frac{1-1}{2}\right)$

Now equate the individual terms to get,

$x=3$

$y=0$

So co-ordinates of Q is (3, 0)

Similarly mid-point R of side AC can be written as,

$R(x, y)=\left(\frac{5-1}{2}, \frac{1+3}{2}\right)$

Now equate the individual terms to get,

$x=2$

 

$y=2$

So co-ordinates of $Q$ is $(2,2)$

Therefore length of median from A to the side BC is,

$\mathrm{AQ}=\sqrt{(-1-3)^{2}+(3-0)^{2}}$

$=\sqrt{16+9}$

$=5$

Similarly length of median from B to the side AC is,

$B R=\sqrt{(1-2)^{2}+(-1-2)^{2}}$

$=\sqrt{1+9}$

$=\sqrt{10}$

Similarly length of median from C to the side AB is

$\mathrm{CP}=\sqrt{(5-0)^{2}+(1-1)^{2}}$

$=\sqrt{25}$

$=5$

 

 

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