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) Given points A(-1,-2), B(1, 0),C(-1, 2) and D(-3,0).
AB = √([1-(-1)]²+[0-(-2)]²) = √(4+4) = √8 = 2√2
BC = √((-1-1)²+(2-0)²) = √(4+4) = √8 = 2√2
CD = √([-3-(-1)]²)+(0-2)²) = √(4+4) = √8 = 2√2
DA = √([-1-(-3)]²)+(-2-0)²) = √(4+4) √8 = 2√2
All the sides of quadrilateral are equal, so it may be a square or rhombus on the basis of its diagonal.
AC = √([1-(-1)]²+[2-(-2)]²) = √(0+16) = 4
BD = √((-3-1)²+(0-0)²) = √(16+0) = 4
Here, AB = BC = CD = DA and AC = BD
Hence, ABCD is a square.

(ii) Given points: A(-3,5), B(3,1), C(0,3) and D(-1,-4).
AB = √([3-(-3)]²+(1-5)²) = √(36+6) = √52 = 2√13
BC = √((0-3)²+(3-1)²) = √(9+4) = √13
CD = √(1-0)²+(-4-3)²) = √(1+49) = √50 = 5√2
DA = √([-3-(-1)]²+[5-(-4)]²) = √(4+81) = √85
AC = √((-1-3)²+(-4-1)²) = √(16+25) = √41
BD = √((-1-3)²+(-4-1)²) = √(16+25) = √41
Here, AC+BC = AB, it means the point C lies on side AB or A,B,C are collinear.
Hence, the quadrilateral ABCD is not possible.

(iii) Given points: A(4,5), B(7,6), C(4,3) and D(1,2).
AB = √(7-4)²+(6-5)²) = √(9+1) = √10
BC = √((4-7)²+(3-6)²) = √(9+9) = √18
CD = √((1-4)²+(2-3)²) = √(9+1) = √10
DA = √((4-1)²+(5-2)]²) = √(9+9) = √18
The opposite sides of quadrilateral are equal. It may be a parallelogram or rectangle. It can be justified with the help of lengths of its diagonal.
AC = √((4-4)²+(3-5)²) = √(0+4) = 2
BD = √((1-7)²+(2-6)²) = √(36+16) = √52 = 2√13
Here, AB = CD, BC = AD and AC ≠ BD.
Hence, ABCD is a parallelogram.

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