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. - Mathematics

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 (2a0) 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 pointsA(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.

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.4 | Q 10 | Page 37