Answer
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Hint: We will find the distance between AC as the length of the diagonal of the square, which is $\sqrt 2 $ times the length of the side of the square. Also, in a square, all the sides are equal. Use the given coordinates to form the equation and find the value of coordinates of B. Use the midpoint of the other coordinates as diagonals of the square bisect each other.
Complete step-by-step answer:
We are given that the opposite vertices of a square are $\left( { - 1,2} \right)$ and $\left( {3,2} \right)$.
Let the coordinates of the vertices of other two opposite sides be $\left( {a,b} \right)$ and $\left( {c,d} \right)$
In a square, all sides are equal and the length of the diagonal is $\sqrt 2 $ times the length of the side of the square.
Also, the distance between two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
We will now calculate the distance between A and C to find the length of the diagonal of the given square.
$
{\text{AC}} = \sqrt {{{\left( { - 1 - 3} \right)}^2} + {{\left( {2 - 2} \right)}^2}} \\
\Rightarrow {\text{AC}} = \sqrt {{4^2} + 0} \\
\Rightarrow {\text{AC}} = 4 \\
$
Then, the length of the side of the square can be calculated by dividing the length of diagonal by $\sqrt 2 $
That is the side of the square is $\dfrac{4}{{\sqrt 2 }} = 2\sqrt 2 $
Since, ABCD is a square, therefore, ${\text{AB = BC = AD = DC}}$
That is the distance between AB is equal to BC.
$\sqrt {{{\left( {a - \left( { - 1} \right)} \right)}^2} + {{\left( {b - 2} \right)}^2}} = \sqrt {{{\left( {3 - a} \right)}^2} + {{\left( {2 - b} \right)}^2}} $
On squaring sides, we will get,
$
{\left( {a + 1} \right)^2} + {\left( {b - 2} \right)^2} = {\left( {3 - a} \right)^2} + {\left( { - 1} \right)^2}{\left( {b - 2} \right)^2} \\
\Rightarrow {\left( {a + 1} \right)^2} = {\left( {3 - a} \right)^2} \\
\Rightarrow {a^2} + 1 + 2a = 9 + {a^2} - 6a \\
\Rightarrow 8a = 8 \\
\Rightarrow a = 1 \\
$
And the length of AB is $2\sqrt 2 $ units.
$
{\text{AB}} = \sqrt {{{\left( { - 1 - a} \right)}^2} + {{\left( {2 - b} \right)}^2}} \\
\Rightarrow 2\sqrt 2 = \sqrt {{{\left( { - 1 - 1} \right)}^2} + {{\left( {2 - b} \right)}^2}} \\
\Rightarrow 2\sqrt 2 = \sqrt {4 + {{\left( {2 - b} \right)}^2}} \\
$
Squaring both sides,
$
8 = 4 + {\left( {2 - b} \right)^2} \\
\Rightarrow 4 = 4 + {b^2} - 4b \\
\Rightarrow {b^2} - 4b = 0 \\
\Rightarrow b\left( {b - 4} \right) = 0 \\
$
Equating each factor to 0, we will get,
$b = 0$
Or
$
b - 4 = 0 \\
b = 4 \\
$
Thus, the coordinates of B are $\left( {1,0} \right)$ and $\left( {1,4} \right)$
It is known that diagonals of a square bisect each other.
Let O be the mid-point of AC and BD.
We will now calculate the coordinates of O using the mid-point formula, when $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ are points, the mid-point is $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Then, the coordinates of O are $\left( {\dfrac{{3 - 1}}{2},\dfrac{{2 + 2}}{2}} \right) = \left( {1,2} \right)$
Point O will also be the mid-point of BD. Let the coordinates of B are $\left( {1,0} \right)$
$\left( {1,2} \right) = \left( {\dfrac{{c + 1}}{2},\dfrac{{d + 0}}{2}} \right)$
On comparing the coordinates we will get,
$
1 = \dfrac{{c + 1}}{2} \\
c + 1 = 2 \\
c = 1 \\
$
And
$
2 = \dfrac{{d + 0}}{2} \\
d = 4 \\
$
Then, the coordinates of D are $\left( {1,4} \right)$
But, if the coordinates of B are $\left( {1,4} \right)$
Then, the coordinates of D are,
$\left( {1,2} \right) = \left( {\dfrac{{c + 1}}{2},\dfrac{{d + 4}}{2}} \right)$
On comparing the coordinates we will get,
$
1 = \dfrac{{c + 1}}{2} \\
c + 1 = 2 \\
c = 1 \\
$
And
$
2 = \dfrac{{d + 4}}{2} \\
d = 0 \\
$
Then, the coordinates of D are \[\left( {1,0} \right)\].
Note: Square is a quadrilateral with 4 equal sides and each angle in a square is ${90^ \circ }$. For this question, one must know the properties of squares and various formulas of coordinate geometry such as distance formula and midpoint formula.
Complete step-by-step answer:
We are given that the opposite vertices of a square are $\left( { - 1,2} \right)$ and $\left( {3,2} \right)$.
Let the coordinates of the vertices of other two opposite sides be $\left( {a,b} \right)$ and $\left( {c,d} \right)$
In a square, all sides are equal and the length of the diagonal is $\sqrt 2 $ times the length of the side of the square.
Also, the distance between two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
We will now calculate the distance between A and C to find the length of the diagonal of the given square.
$
{\text{AC}} = \sqrt {{{\left( { - 1 - 3} \right)}^2} + {{\left( {2 - 2} \right)}^2}} \\
\Rightarrow {\text{AC}} = \sqrt {{4^2} + 0} \\
\Rightarrow {\text{AC}} = 4 \\
$
Then, the length of the side of the square can be calculated by dividing the length of diagonal by $\sqrt 2 $
That is the side of the square is $\dfrac{4}{{\sqrt 2 }} = 2\sqrt 2 $
Since, ABCD is a square, therefore, ${\text{AB = BC = AD = DC}}$
That is the distance between AB is equal to BC.
$\sqrt {{{\left( {a - \left( { - 1} \right)} \right)}^2} + {{\left( {b - 2} \right)}^2}} = \sqrt {{{\left( {3 - a} \right)}^2} + {{\left( {2 - b} \right)}^2}} $
On squaring sides, we will get,
$
{\left( {a + 1} \right)^2} + {\left( {b - 2} \right)^2} = {\left( {3 - a} \right)^2} + {\left( { - 1} \right)^2}{\left( {b - 2} \right)^2} \\
\Rightarrow {\left( {a + 1} \right)^2} = {\left( {3 - a} \right)^2} \\
\Rightarrow {a^2} + 1 + 2a = 9 + {a^2} - 6a \\
\Rightarrow 8a = 8 \\
\Rightarrow a = 1 \\
$
And the length of AB is $2\sqrt 2 $ units.
$
{\text{AB}} = \sqrt {{{\left( { - 1 - a} \right)}^2} + {{\left( {2 - b} \right)}^2}} \\
\Rightarrow 2\sqrt 2 = \sqrt {{{\left( { - 1 - 1} \right)}^2} + {{\left( {2 - b} \right)}^2}} \\
\Rightarrow 2\sqrt 2 = \sqrt {4 + {{\left( {2 - b} \right)}^2}} \\
$
Squaring both sides,
$
8 = 4 + {\left( {2 - b} \right)^2} \\
\Rightarrow 4 = 4 + {b^2} - 4b \\
\Rightarrow {b^2} - 4b = 0 \\
\Rightarrow b\left( {b - 4} \right) = 0 \\
$
Equating each factor to 0, we will get,
$b = 0$
Or
$
b - 4 = 0 \\
b = 4 \\
$
Thus, the coordinates of B are $\left( {1,0} \right)$ and $\left( {1,4} \right)$
It is known that diagonals of a square bisect each other.
Let O be the mid-point of AC and BD.
We will now calculate the coordinates of O using the mid-point formula, when $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ are points, the mid-point is $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Then, the coordinates of O are $\left( {\dfrac{{3 - 1}}{2},\dfrac{{2 + 2}}{2}} \right) = \left( {1,2} \right)$
Point O will also be the mid-point of BD. Let the coordinates of B are $\left( {1,0} \right)$
$\left( {1,2} \right) = \left( {\dfrac{{c + 1}}{2},\dfrac{{d + 0}}{2}} \right)$
On comparing the coordinates we will get,
$
1 = \dfrac{{c + 1}}{2} \\
c + 1 = 2 \\
c = 1 \\
$
And
$
2 = \dfrac{{d + 0}}{2} \\
d = 4 \\
$
Then, the coordinates of D are $\left( {1,4} \right)$
But, if the coordinates of B are $\left( {1,4} \right)$
Then, the coordinates of D are,
$\left( {1,2} \right) = \left( {\dfrac{{c + 1}}{2},\dfrac{{d + 4}}{2}} \right)$
On comparing the coordinates we will get,
$
1 = \dfrac{{c + 1}}{2} \\
c + 1 = 2 \\
c = 1 \\
$
And
$
2 = \dfrac{{d + 4}}{2} \\
d = 0 \\
$
Then, the coordinates of D are \[\left( {1,0} \right)\].
Note: Square is a quadrilateral with 4 equal sides and each angle in a square is ${90^ \circ }$. For this question, one must know the properties of squares and various formulas of coordinate geometry such as distance formula and midpoint formula.
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