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 ${\displaystyle \left(x_1{,}_{y}1\right)}$ and ${\displaystyle (x_2,y_2)}$ is given by the formula

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

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.

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

${\displaystyle =\sqrt{{(-1)}^2+{(-5)}^2}}$

${\displaystyle =\sqrt{1+25}}$

${\displaystyle AB=\sqrt{26}}$

${\displaystyle BC=\sqrt{{(3+2)}^2+{(4-3)}^2}}$

${\displaystyle =\sqrt{{\left(5\right)}^2+{\left(1\right)}^2}}$

${\displaystyle =\sqrt{25+1}}$

${\displaystyle BC=\sqrt{26}}$

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

${\displaystyle =\sqrt{{(-5)}^2+{(-1)}^2}}$

${\displaystyle =\sqrt{25+1}}$

${\displaystyle CD=\sqrt{26}}$

${\displaystyle AD=\sqrt{{(2+3)}^2+{(-1+2)}^2}}$

${\displaystyle =\sqrt{{\left(5\right)}^2+{\left(1\right)}^2}}$

${\displaystyle =\sqrt{25+1}}$

${\displaystyle AD=\sqrt{26}}$

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