Question

$\left(0,-1\right)$ and $\left(0,3\right)$ are two opposite vertices of a square. The other two vertices are

• A. $\left(0,1\right),\left(0,-3\right)$
• B. $\left(3,-1\right),\left(0,0\right)$
• C. $\left(2,1\right),\left(-2,1\right)$
• D. $\left(2,2\right),\left(1,1\right)$

Answer 1

The correct option is C

$\left(2,1\right),\left(-2,1\right)$

Explanation for the correct option:

Step 1. Draw a square $ABCD$ with the diagonals $AC$ and $BD$ which are intersecting at $E$

Let $A(0,–1)$ and $C(0,3)$ be its two opposite vertices.

Step 2: Find the diagonal $AC$ and length of each side

Length of diagonal $\begin{array}{rcl}AC& =& \sqrt{{(0-0)}^2+{(3+1)}^2}\\ & =& \sqrt{16}\\ & =& 4\end{array}$

Length of each side $=\frac{\mathbf{d}\mathbf{i}\mathbf{a}\mathbf{g}\mathbf{o}\mathbf{n}\mathbf{a}\mathbf{l}}{\sqrt{\mathbf{2}}}$

$$\begin{gathered} =\frac{4}{\sqrt{2}} \\ =2\sqrt{2} \end{gathered}$$

Now, Let the co-ordinates of another vertex of the square $B$ be $(x_1,y_1)$

Step 3: Find the $y$ coordinate

The distance of vertex $B$from vertex $A$and vertex $C$will be equal.

$\sqrt{{(x_1-0)}^2+{(y_1+1)}^2}=\sqrt{{(x_1-0)}^2+{(y_1-3)}^2}$

$\Rightarrow$${x_1}^2+{y_1}^2+2y_1+1={x_1}^2+{y_1}^2–6y_1+9$

$\Rightarrow$ $8y_1=8$

$\Rightarrow$ $y_1=1$

Step 4: Find the $x$ coordinate

Distance $BA=2\sqrt{2}$

$\Rightarrow$${(x_1-0)}^2+{(y_1+1)}^2=2\sqrt{2}$

$\Rightarrow$ ${x_1}^2+{y_1}^2+2y_1+1=8$

$\Rightarrow$ ${x_1}^2+4=8$ $\left[\because y_1=1\right]$

$\Rightarrow$ ${x_1}^2=4$

$\Rightarrow$ $x_1=\pm 2$

Therefore, The required two vertices of the square $ABCD$ are $(+2,1)$ and $(–2,1)$.

Hence, option ‘C’ is Correct.