#### 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 `(x_1,y_1)` and `(x_2, y_2)` is given by the formula

`d = sqrt((x_1-x_2)^2 + (y_1 - y_2)^2)`

The co-ordinates of the midpoint `(x_m,y_m)` between two points `(x_1, y_1)` and `(x_2, y_2)` is given by,

`(x_m,y_m) = (((x_1 + x_2)/2)"," ((y_1+y_2)/2))`

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.

`(x_D,y_D) = (((1 + 5)/2)"," ((-1+1)/2))`

`(x_D, y_D) = (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*’.

`AD = sqrt((-1-3)^2 + (3 - 0)^2)`

`= sqrt((-4)^2 + (3)^2)`

`= sqrt(16 + 9)`

AD = 5

Thus the length of the median through the vertex ‘*A*’ of the given triangle is 5 units