Answer :

**To Find: Type of quadrilateral formed**

(i) Let the points ( - 1, - 2), (1, 0), ( - 1, 2), and ( - 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x_{1}, y_{1}) and (x_{2}, y_{2}).

_{2}- x

_{1})

^{2}+ (y

_{2}- y

_{1})

^{2}

AB = [(-1- 1)^{2} + (-2- 0)^{2}]^{1/2}

= =

= 2√2

BC = √[(1 + 1)^{2} + (0- 2)^{2}]

= =

= 2√2

CD = √[(-1 + 3)^{2} + (2- 0)^{2}]

= =

= 2√2

AD = √[(-1+ 3)^{2} + (-2- 0)^{2}]

= =

= 2√2

Diagonal AC = √[(-1 + 1)^{2} + (-2 - 2)^{2}]

=

= 4

Diagonal BD = √[(1 + 3)^{2} + (0- 0)^{2}]

=

= 4

It is clear that all sides of this quadrilateral are of the same length and the diagonals are of the same length. Therefore, the given points are the vertices of a square

(ii)Let the points (- 3, 5), (3, 1), (0, 3), and ( - 1, - 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x_{1}, y_{1}) and (x_{2}, y_{2}).

_{2}- x

_{1})

^{2}+ (y

_{2}- y

_{1})

^{2}

AB = √[(-3- 3)^{2} + (5- 1)^{2}]

= =

= 2√13

BC = √[(3- 0)^{2} + (1- 3)^{2}]

=

=

CD = √[(0 + 1)^{2} + (3+ 4)^{2}]

= =

= 5√2

AD = √[(-3+ 1)^{2} + (5+ 4)^{2}]

=

=

We can observe that all sides of this quadrilateral are of different lengths.

Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc

(iii)Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x_{1}, y_{1}) and (x_{2}, y_{2}).

_{2}- x

_{1})

^{2}+ (y

_{2}- y

_{1})

^{2}

AB = √[(4- 7)^{2} + (5 - 6)^{2}]

=

=

BC = √[(7- 4)^{2} + (6 - 3)^{2}]

=

=

CD = √[(4- 1)^{2} + (3 - 2)^{2}]

=

=

AD = √[(4- 1)^{2} + (5 - 2)^{2}]

=

=

Diagonal AC =√[(4 - 4)^{2} + (5- 3)^{2}]

=

= 2

Diagonal BD = √[(7 - 1)^{2} + (6 - 2)^{2}]

=

=

= 2√13

We can observe that opposite sides of this quadrilateral are of the same length.

However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram

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