Question

prove that the diagonals of a rectangle ABCD,with vertices A(2,-1),B(5,-1),C( 5,6) and D(2,6) are equal and bisect each other .

Answer 1

One diagonal AC isroot 55 and another BD isroot 40 ithink its correct

Answer 2

Answer:

The diagonals of a rectangle are equal and bisect each other.

Step-by-step explanation:

The vertices of rectangle are A(2,-1),B(5,-1),C( 5,6) and D(2,6).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The diagonal are AC and BD.

$AC=\sqrt{(5-2)^2+(6+1)^2}=\sqrt{9+49}=\sqrt{58}$

$BD=\sqrt{(2-5)^2+(6+1)^2}=\sqrt{9+49}=\sqrt{58}$

The length of both diagonals are same.

Midpoint of both diagonals,

$\text {Midpoint of AC}=(\frac{2+5}{2},\frac{-1+6}{2})=(\frac{7}{2},\frac{5}{2})$

$\text {Midpoint of BD}=(\frac{5+2}{2},\frac{-1+6}{2})=(\frac{7}{2},\frac{5}{2})$

The midpoint of both diagonals is $(\frac{7}{2},\frac{5}{2})$. It means both diagonal bisects each other.

Hence proved that the diagonals of a rectangle are equal and bisect each other.