Two opposite vertices of a square are (−1, 2) and (3, 2).

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

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  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.

$A C=\sqrt{(-1-3)^{2}+(2-2)^{2}}$

$=\sqrt{(-4)^{2}+(0)^{2}}$

$=\sqrt{16}$

$A C=4$

Now we know that in a square,

Side of the square $=\frac{\text { Diagonal of the square }}{\sqrt{2}}$

Substituting the value of the diagonal we found out earlier in this equation we have,

Side of the square $=\frac{4}{\sqrt{2}}$

Side of the square $=2 \sqrt{2}$

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’.

$A P=\sqrt{(-1-x)^{2}+(2-y)^{2}}$

$C P=\sqrt{(3-x)^{2}+(2-y)^{2}}$

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

$A P=C P$

$\sqrt{(-1-x)^{2}+(2-y)^{2}}=\sqrt{(3-x)^{2}+(2-y)^{2}}$

Squaring on both sides we have,

$(-1-x)^{2}+(2-y)^{2}=(3-x)^{2}+(2-y)^{2}$

$1+x^{2}+2 x+4+y^{2}-4 y=9+x^{2}-6 x+4+y^{2}-4 y$

$8 x=8$

$x=1$

Substituting this value of ‘x’ and the length of the side in the equation for ‘AP’ we have,

$A P=\sqrt{(-1-x)^{2}+(2-y)^{2}}$

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

 

$2 \sqrt{2}=\sqrt{(-2)^{2}+(2-y)^{2}}$

Squaring on both sides,

$8=(-2)^{2}+(2-y)^{2}$

 

$8=4+4+y^{2}-4 y$

$0=y^{2}-4 y$

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

$y^{2}-4 y=0$

$y(y-4)=0$

The roots of this equation are 0 and 4.

Therefore the other two vertices of the square are.