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 (0,2b); A (2a, 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.
