Question

Show that the points (–4, –1), (–2, –4), (4, 0) and (2, 3) are the vertices of a rectangle.

Answer 1

The distance d between two points (x₁, y₁) and (x₂, y₂) is given by the formula

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

In a rectangle, the opposite sides are equal in length. The diagonals of a rectangle are also equal in length.

Here the four points are A(−4,−1), B(−2,−4), C(4, 0) and D(2, 3).

First let us check the length of the opposite sides of the quadrilateral that is formed by these points.

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

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

`= sqrt(4 + 9)`

`AB = sqrt(13)`

`CD = sqrt((4 - 2)^2 + (0 -3)^2)`

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

`= sqrt(4+9)`

`CD = sqrt(13)`

We have one pair of opposite sides equal.

Now, let us check the other pair of opposite sides.

`BC = sqrt((-2-4)^2+(-4-0)^2)`

`=sqrt((-6)^2 + (-4)^2)` 

`=sqrt(36 + 16)`

`BC = sqrt(52) `

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

`= sqrt((-6)^2 + (-4)^2)` 

`=sqrt(36 + 16) `

`BC = sqrt(52)`

The other pair of opposite sides are also equal. So, the quadrilateral formed by these four points is definitely a parallelogram.

For a parallelogram to be a rectangle we need to check if the diagonals are also equal in length.

`AC = sqrt((-4-4)^2 + (-1-0)^2)`

`= sqrt((-8)^2 + (-1)^2)`

`= sqrt(64+1)`

`AC = sqrt(65)`

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

`= sqrt((-4)^2 + (-7)^2)`

`= sqrt(16+49)`

`BD = sqrt(65)`

Now since the diagonals are also equal we can say that the parallelogram is definitely a rectangle.

Hence we have proved that the quadrilateral formed by the four given points is a rectangle.