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)

Lets identify the types of Quadrilaterals based on their lengths of sides and their lengths of diagonals.
We know that, the distance between two points A(x1,y1), B(x2,y2) is √(x2-x1)²+(y2-y1)²
(i)(-1,-2),(1,0),(-1,2),(-3,0)
Let the given points are A(-1,-2),B(1,0),C(-1,2) and D(-3,0)
Then, AB=√(1+1)²+(0+2)²=√2²+2²=√4+4=√8
BC=√(-1-1)²+(2-0)²=√(-2)²+2²=√4+4=√8
CD=√((-3)-(-1))²+(0-2)²=√(-2)²+(-2)²=√4+4=√8
DA=√(-3-(-1))²+(0-(-2))²=√(-2)²+2²=√4+4=√8
AC=√((-1)-(-1))²+(2-(-2))²=√0+4²=√16=4
BD=√(-3-1)²+(0-0)²=√(-4)²=√16=4
Since the four sides AB,BC,CD and DA are equal and the diagonals AC and BD are equal.
∴Quadrilateral ABCD is a square
(ii)(-3,5),(3,1),(0,3),(-1,-4)
Let the given points are A(-3,5),B(3,1),C(0,3) and D(-1,-4)
Then AB=√(-3-3)²+(5-1)²=√(-6)²+4²=√36+16=√52
BC=√(3-0)²+(1-3)²=√(3²+(-2)²)=√9+4=√13
CD=√(0-(-1))²+(3-(-4))²=√1²+(7)²=√1+49=√50
DA=√(-1-(-3))²+(-4-5)²=√(2)²+(-9)²=√4+81=√85
Here AB≠BC≠CD≠DA ∴it is a quadrilateral
(iii)(4,5),(7,6),(4,3),(1,2)
Let the given points are A(4,5),B(7,6),C(4,3) and D(1,2)
Then AB=√(7-4)²+(6-5)²=√3²+1²=√9+1=√10
BC=√(4-7)²+(3-6)²=√(-3)²+(-3)²=√9+9=√18
CD=√(1-4)²+(2-3)²=√(-3)²+(-1)²=√9+1=√10
DA=√(1-4)²+(2-5)²=√(-3)²+(-3)²=√9+9=√18
AC=√(4-4)²+(3-5)²=√0+(-2)²=√4=2
BD=√(1-7)²+(2-6)²=√(-6)²+(-4)²=√36+16=√52
Here AB=CD,BC=DA. But AC≠BD
Hence the pairs of opposite sides are equal but diagonal are not equal so it is a parallelogram.