Question

Using coordinate geometry, prove that the diagonals of a rectangle bisect each other and are equal

Answer 1

Put the lower left corner at the origin A(0,0) and let the other three corners be B(0,B),C(a,b),D(a,0). Label as indicated.

Equality:

AC²:(a-0)²+(b-0)²=a²+b²
BD²:(0-a)²+(b-0)²=a²+b²

Thus AC=BD

Bisection.Compute the midpoints of each and show it is the same point for each diagonal.

midpoint of AC:(0+a)/2,(0+b)/2=(a/2,b/2)

midpoint of BD:(o+a)/2,(b+0)/2=(a/2,b/2)

Since the midpoints are common,that is where they cross. since where they cross are the midpoints, the diagonals bisect each other.

HENCE PROVED!! 

Answer 2

Answer:

Let OABC be a rectangle such that OA is along x axis and OB is along y axis

also, let OA be a and OB be b

Therefore coordinates of A are (a,0) and that of B are (b,0)

We have OABC is a rectangle

therefore AC = OB 

i.e AC =b

similarly,

OA= a

Therefore coordinates of the midpoint of OC are (a/2, b/2)

similarly, midpoints of AB are (a/2,b/2)

since midpoints are same,

therefore OC = AB

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