Question

Show that the quadrilateral whose vertices are (2, −1), (3, 4) (−2, 3) and (−3,−2) is a rhombus.

Answer 1

The distance d between two points `(x_1,_y1)` and `(x_2, y_2)` is given by the formula

`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`

In a rhombus all the sides are equal in length.

Here the four points are A (2, −1), B (3,  4), C (−2, 3) and D (−3, −2).

First let us check if all the four sides are equal.

`AB = sqrt((2 - 3)^2 + (-1 - 4)^2)`

`= sqrt((-1)^2 + (-5)^2)`

`= sqrt(1 + 25)`

`AB = sqrt26`

`BC = sqrt((3 + 2)^2 + (4 - 3)^2)`

`= sqrt((5)^2 + (1)^2)`

`= sqrt(25 + 1)`

`BC = sqrt26`

CD = sqrt((3 + 2)^2 + (4 - 3)^2)

`= sqrt((-5)^2 + (-1)^2)`

`=sqrt(25 + 1)`

`CD = sqrt26`

`AD = sqrt((2 + 3)^2 + (-1 + 2)^2)`

`= sqrt((5)^2 + (1)^2)`

`= sqrt(25 + 1)`

`AD = sqrt26`

Here, we see that all the sides are equal, so it has to be a rhombus.