If A (−1, 3), B (1, −1) and C (5, 1) are the vertices
Question:

If A (−1, 3), B (1, −1) and C (5, 1) are the vertices of a triangle ABC, find the length of the median through A.

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

The co-ordinates of the midpoint $\left(x_{n}, y_{n}\right)$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by,

$\left(x_{m}, y_{m}\right)=\left(\left(\frac{x_{1}+x_{2}}{2}\right),\left(\frac{y_{1}+y_{2}}{2}\right)\right)$

Here, it is given that the three vertices of a triangle are A(−1,3), B(1,−1) and C(5,1).

The median of a triangle is the line joining a vertex of a triangle to the mid-point of the side opposite this vertex.

Let ‘D’ be the mid-point of the side ‘BC’.

Let us now find its co-ordinates.

$\left(x_{D}, y_{D}\right)=\left(\left(\frac{1+5}{2}\right),\left(\frac{-1+1}{2}\right)\right)$

$\left(x_{D}, y_{D}\right)=(3,0)$

Thus we have the co-ordinates of the point as D(3,0).

Now, let us find the length of the median ‘AD’.

$A D=\sqrt{(-1-3)^{2}+(3-0)^{2}}$

$=\sqrt{(-4)^{2}+(3)^{2}}$

$=\sqrt{16+9}$

$A D=5$

Thus the length of the median through the vertex ‘ $A$ ‘ of the given triangle is 5 units.