Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

i)  Let A (- 1, - 2), B (1, 0), C (- 1, 2), and D (- 3, 0) be the four points of the quadrilateral.

To find AB, that is, distance between points A (- 1, - 2) and B (1, 0) by using the distance formula,

AB = √(1 + 1)² + (0 + 2)²

= √(2)² + (2)²

= √4 + 4

= √8

= 2√2 units

To find BC, that is, distance between points B (1, 0) and C (- 1, 2) by using the distance formula,

BC = √ (- 1 - 1)² + (2 - 0)²

= √ (- 2)² + 2²

= √4 + 4

= √8

= 2√ 2 units

To find CD, that is, distance between points C (- 1, 2), and D (- 3, 0) by using the distance formula,

CD = √(- 3 + 1)² + (0 - 2)²

= √ (- 2)² + (- 2)²

= √4 + 4

= √8

= 2√2 units

To find AD, that is, distance between Points A (- 1, - 2) and D (- 3, 0) using distance formula,

AD = √(- 3 + 1 )² + (0 + 2)²

= √ (- 2 )² + (2)²

= √4 + 4

= √ 8

= 2√2 units

To find AC, the  diagonals of the quadrilateral , that is, distance between points A (- 1, - 2) and C (- 1, 2)

Diagonal AC = √( - 1 + 1)² + (2 + 2)²

= √0² + 4²

= 4 units

To find BD, that is, distance between points B (1, 0) and D (- 3, 0)

Diagonal BD = √( - 3 - 1 )² + (0 + 0)²

= √ (- 4)² + 0²

= 4 units

The four sides AB, BC, CD, and AD are of the same length, and diagonals AC and BD are of equal length. Therefore, ABCD is a square.

ii) Let A (- 3, 5), B (3, 1), C (0, 3), and D (- 1, - 4) be the four points of the quadrilateral.

We know that the distance between any two points is given by the distance formula = √(x₁ - x₂)² + (y₁ - y₂)² 

To find AB, that is, the distance between points A (- 3, 5) and B (3, 1) by using the distance formula,

AB = √(3 + 3)² + (1 - 5)²

= √(6)² + (- 4)²

= √36 + 16

= √52

= 2√13 units

To find BC, that is, distance between points B (3, 1) and C (0, 3) by using the distance formula,

BC = √(0 - 3)² + (3 - 1)²

= √ (-3)² + (2)²

= √9 + 4

= √13 units

To find CD, that is, distance between Points C (0, 3), and D (- 1, - 4) by using the distance formula,

CD = √(- 1 - 0)² + (- 4 - 3)²

= √ (- 1)² + (- 7)²

= √1 +  49

= √50

= 5√2 units

To find AD, that is, distance between Points A (- 3, 5) and D (- 1, - 4) using distance formula,

AD = √(-1 + 3)² + (- 4 - 5)²

= √ (2)² + (- 9)²

= √4 + 81

= √85 units

Since, AB ≠ BC ≠ CD ≠ AD, therefore, no special quadrilateral can be formed from the given vertices.

iii) Let A (4, 5), B (7, 6), C (4, 3) and D (1, 2) be the four points of the quadrilateral.

We know that the distance between any two points is given by the Distance Formula,

Distance Formula = √ (x₁ - x₂)² + (y₁ - y₂)²  

To find AB, that is, distance between points A (4, 5) and B (7, 6), by using the distance formula,

AB = √(7 - 4)² + (6 - 5)²

= √3² + 1²

= √9 + 1

= √10 units

To find BC, that is, distance between points B (7, 6) and C (4, 3) by using the distance formula,

BC = √ (4 - 7 )² + (3 - 6)²

= √ (- 3)² + 3²

= √9 + 9

= √18 units

To find CD, that is, distance between points C (4, 3) and D (1, 2) by using the distance formula,

CD = √(1 - 4)² + (2 - 3)²

= √ (- 3)² + (- 1)²

= √9 + 1

= √10 units

To find AD i.e. Distance between points A (4, 5) and D (1, 2) using distance formula,

AD = √( 1 - 4 )² + ( 2 - 5)²

= √ (- 3)² + (- 3)²

= √ 9 + 9

= √18 units

To find AC, the distance between points A (4, 5) and C (4, 3), we have

Diagonal AC = √( 4 - 4 )² + ( 3 - 5)²

= √( 0 )² + ( - 2)²

= 2 units

To find BD, distance between points B (7, 6) and D (1, 2), we have

Diagonal BD = √(1 - 7)² + (2 - 6)²

= √ ( - 6 )² + ( - 4 )²

= √ 36 + 16

= √52 units

Since AB = CD and BC = AD, but the diagonals AC ≠  BD, thus the quadrilateral is a parallelogram.