Question

Prove that the points A(1, 7), B(4, 2), C(−1, −1), and D(−4, 4) are the vertices of a square.

Answer 1

The distance d between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula,

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

In a square, all the sides are equal in length. Also, the diagonals are equal in length in a square.

Here the four points are A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4).

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

`AB = sqrt((1- 4)^2 + (7 - 2)^2)`

`= sqrt((-3)^2 + (5)^2)`

`= sqrt(9 + 25)`

`BC = sqrt34`

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

`= sqrt((-5)^2 + (-3)^2 )`

`= sqrt(25 + 9)`

`CD = sqrt(34)`

`AD = sqrt((1 +4)^2 + (7 - 4)^2)`

`= sqrt((5)^2 + (3)^2)`

`= sqrt(25 + 9)`

`AD = sqrt34`

Since all the sides of the quadrilateral are the same, it is a rhombus.

For the rhombus to be a square, the diagonals must also be equal.

`AC = sqrt((1 + 1)^2 + (7 + 1)^2)`

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

`= sqrt(4 + 64)`

`AC = sqrt(68)`

`BD = sqrt((4 + 4)^2 + [2 +(-4)]^2)`

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

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

`= sqrt(64 + 4)`

`BD = sqrt(68)`

Since the diagonals of the rhombus are equal, the rhombus is a square.

Hence, the quadrilateral formed by the given points is a square.