Prove that $(0, 5) (- 2, 2), (5, 0)$ and $(7, 7)$ are the vertices of a rhombus

Open in App

Solution

$A (0, 5), B (- 2, - 2), C (5, 0), D (7, 7)$ be vertices of quadrilateral.
$A B = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$
$= \sqrt{(0 + 2)^{2} + (5 + 2)^{2}}$
$= \sqrt{4 + 49} = \sqrt{53}$
$B C = \sqrt{(- 2 - 5)^{2} + (- 2 - 0)^{2}} B (- 2, 2); C (5, 0)$
$= \sqrt{49 + 4} = \sqrt{53}$
$C D = \sqrt{(5 - 7)^{2} + (0 - 7)^{2}} C (5, 0); D (7, 7)$
$\sqrt{4 + 49} = \sqrt{53}$
$A D = \sqrt{(0 - 7)^{2} + (5 - 7)^{2}} A (0, 5); D (7, 7)$
$= \sqrt{49 + 4} = \sqrt{53}$
$A C = \sqrt{(0 - 5)^{2} + (5 - 0)^{2}} A (0, 5); C (5, 0)$
$= \sqrt{25 + 25} = \sqrt{50}$
$B D = \sqrt{(- 2 - 7)^{2} + (- 2 - 7)^{2}} B (- 2, - 2); D (7, 7)$
$= \sqrt{81 + 81} = \sqrt{162}$
$A B = B C = C D = D A$ and $A C \neq B D$
The given point form a Rhombus.

Suggest Corrections

Similar questions

Prove that the points (0, 5), B(-2, -2), C(5, 1) and D(7, 7) are the vertices of a rhombus.

(4, - 1), (6, 0), (7, 2)

(5, 1)

(- 3, 2), (- 5, - 5), (2, - 3)

(4, 4)

If A(3,-4),B(4,5),C(5,-5),D(4,-10) are the vertices of a quadrilateral prove that ABCD is Rhombus?

Join BYJU'S Learning Program