Question

Show that the points A(2, -1), B (3, 4) , C (-2, 3) and D (-3, -2) are the vertices of a rhombus.​

Answer 1

Answer:

Let the point A(2,-1), B(3,4), C(-2,3) and D(-3,-2) be the vertices of a rhombus ABCD.

We know that all the sides of a rhombus ABCD are equal.

So by distance formula,

Distance between two points

AB = $\sqrt{((x2-x1)^{2} +(y2-y1)^{2}}$

AB= $\sqrt{(3-2)^{2}+(4+1)^{2} } =\sqrt{1^{2} +5^{2} } =\sqrt{26}$

BC=$\sqrt{(-2-3)^{2}+(3-4)^{2} } =\sqrt{-5^{2}+ -1^{2} } =\sqrt{26}$

use for all points u get $\sqrt{26}$ as answer.

∴AB=BC=CD=AD

Hence, ABCD is a rhombus (Proved)