In Fig. 14.36, a right triangle BOA is given C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A and B.

We have a right angled triangle,`triangle BOA` right angled at O. Co-ordinates are B (0,2*b*); A (2*a**, *0) and C (0, 0).

#### Solution

We have to prove that mid-point C of hypotenuse AB is equidistant from the vertices.

In general to find the mid-pointP(x,y) of two points`A(x_1,y_1)`and `B (x_2,y_2)` we use section formula as,

`p(x,y)=((x_1+x_2)/2,(y_1+y_2)/2)`

So co-rdinates of C is ,

C (a,b)

In general, the distance between` A(x_1,y_2)` and `B(x_2,y_2)`is given by,

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

So,

`CO=sqrt((a-0)^2+(b0o)^2)`

`=sqrt(a^2+b^2)`

`CB =sqrt((a-0)^2+(b-2b)^2)`

`sqrt(a^2+b^2)`

`CA=sqrt((a-2a)^2+(b-0)^2)

`sqrt(a^2+b^2`

Hence, mid-point C of hypotenuse AB is equidistant from the vertices.