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


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


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


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.

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



$A C=4$

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.

$A P=C P$


Squaring on both sides we have,


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

$8 x=8$


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,



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

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

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

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


The roots of this equation are 0 and 4.

Therefore the other two vertices of the square are.


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