Find the length of the medians of a ΔABC having
Question:

Find the length of the medians of a ΔABC having vertices at A(0, −1), B(2, 1) and C(0, 3).

Solution:

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (0,−1); B (2, 1) and C (0, 3).

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,

$P(x, y)=\left(\frac{2+0}{2}, \frac{1-1}{2}\right)$

Now equate the individual terms to get,

$x=1$

 

$y=0$

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

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

$Q(x, y)=\left(\frac{2+0}{2}, \frac{3+1}{2}\right)$

Now equate the individual terms to get,

$x=1$

 

$y=2$

So co-ordinates of Q is (1, 2)

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

$R(x, y)=\left(\frac{0+0}{2}, \frac{3-1}{2}\right)$

Now equate the individual terms to get,

$x=0$

 

$y=1$

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

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

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

$=\sqrt{1+9}$

 

$=\sqrt{10}$

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

BR $=\sqrt{(2-0)^{2}+(1-1)^{2}}$

$=\sqrt{4}$

 

$=2$

Similarly length of median from C to the side AB is

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

$=\sqrt{1+9}$

$=\sqrt{10}$

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.