Question

prove that the diagonals of a rectangle bisect each other and are equal |coordinate geometry | class 10 |

Answer 1

Proof:
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 mid point of OC are (a/2, b/2)

similarly mid points of AB are (a/2,b/2)

since mid points are same,
therefore OC = AB

Hence proved

Answer 2

let OABC be a rectangle

O(0,0)

A(a,0)

B(a,b)

C(0,b)

midpoint of OB=(x1+x2/2 , y1+y2/2)

                         =(0+a/2 , 0+b/2)

                         =(a/2 , b/2)

similarly, AC=(a/2 , b/2)

  since the midpoints of the diagonal coincides with each other ,the diagonal bisects each other

OB=\sqrt(a-0)^2+(b-0)^=\sqrt(a^2+b^2)

AC=\sqrt(a-0)^2+(b-0)^=\sqrt(a^2+b^2)

∴OB=AC

∴the diagonals bisect each other and are equal.