Question

Two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of other two
vertices.

Answer 1

The distance d between two points `(x_1,y_1)` and `(x_2,y_2)`

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

In a square, all the sides are of equal length. The diagonals are also equal to each other. Also in a square, the diagonal is equal to `sqrt2` times the side of the square.

Here let the two points which are said to be the opposite vertices of a diagonal of a square be A(−1,2) and C(3,2).

Let us find the distance between them which is the length of the diagonal of the square.

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

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

`= sqrt(16)`

AC = 4

Now we know that in a square,

The side of the square = `"Diagonal of the square"/sqrt2`

The side of the square = `2sqrt2`

Now, a vertex of a square has to be at equal distances from each of its adjacent vertices.

Let P(x, y) represent another vertex of the same square adjacent to both ‘A’ and ‘C’

`AP = sqrt((-1-x)^2 + (2 -y)^2)`

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

But these two are nothing but the sides of the square and need to be equal to each other.

AP = CP

`sqrt((-1-x)^2 + (2 - y)^2) = sqrt((3 - x)^2 + (2 - y)^2)`

Squaring on both sides we have,

`AP = sqrt((-1-x)^2 + (2 - y)^2)`

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

`2sqrt2 = sqrt((-2)^2 + (2 - y)^2)`

Squaring on both sides,

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

`8 = 4 + 4 = y^2 - 4y`

`0 = y^2 -   4y`

We have a quadratic equation. Solving for the roots of the equation we have,

`y^2 - 4y = 0`

y(y - 4) = 0

The roots of this equation are 0 and 4.

Therefore the other two vertices of the square are (1, 0) and (1,4)