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)
Using Distance Formula to find distances AB, BC, CD and DA, we get

AB = \(\sqrt{(1 + 1)^2 + (0 + 2)^2} \)
\( = \sqrt{4 + 4} \)
\(= 2 \sqrt{2}\)

BC = \(\sqrt{(-1 - 1)^2 + (2 - 0)^2} \)
\( = \sqrt{4 + 4} \)
\( = 2 \sqrt{2}\)

CD = \(\sqrt{(-3 + 1)^2 + (0 - 2)^2} \)
\( = \sqrt{4 + 4} \)
\( = 2 \sqrt{2}\)

DA = \(\sqrt{(-3 + 1)^2 + (0 - 2)^2} \)
\( = \sqrt{4 + 4} \)
\( = 2 \sqrt{2}\)

Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.

AC = \(\sqrt{(-1 + 1)^2 + (2 + 2)^2} \)
\( = \sqrt{0 + 16} \)
\( = 4\)

BD = \(\sqrt{(-3 - 1)^2 + (0 + 0)^2} \)
\( = \sqrt{16 + 0} \)
\( = 4\)

Therefore, diagonals of quadrilateral ABCD are also equal. … (2)

From (1) and (2), we can say that ABCD is a square.

(ii) Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = \(\sqrt{(-3 - 3)^2 + (1 - 5)^2} \)
\( = \sqrt{36 + 16} \)
\( = 2 \sqrt{13}\)

BC = \(\sqrt{(0 - 3)^2 + (3 - 1)^2} \)
\( = \sqrt{9 + 4} \)
\( = \sqrt{13}\)

CD = \(\sqrt{(-1 - 0)^2 + (-4 - 3)^2} \)
\( = \sqrt{1 + 49} \)
\( = 5\sqrt{2}\)

DA = \(\sqrt{(-1 + 3)^2 + (-4 - 5)^2} \)
\( = \sqrt{4 + 81} \)
\( = \sqrt{85}\)

We cannot find any relation between the lengths of different sides.

Therefore, we cannot give any name to the figure ABCD.

(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get

AB = \(\sqrt{(7 - 4)^2 + (6 - 5)^2} \)
\( = \sqrt{9 + 1} \)
\( = \sqrt{10}\)

BC = \(\sqrt{(4 - 7)^2 + (3 - 6)^2} \)
\( = \sqrt{9 + 9} \)
\( = \sqrt{18}\)

CD = \(\sqrt{(1 - 4)^2 + (2 - 3)^2} \)
\( = \sqrt{9 + 1} \)
\( = \sqrt{10}\)

DA = \(\sqrt{(1 - 4)^2 + (2 - 5)^2} \)
\( = \sqrt{9 + 9} \)
\( = \sqrt{18}\)

Here opposite sides of quadrilateral ABCD are equal. … (1)

We can now find out the lengths of diagonals.

AC = \(\sqrt{(4 - 4)^2 + (3 - 5)^2} \)
\( = \sqrt{0 + 4} \)
\( = 2 \)

BD = \(\sqrt{(1 - 7)^2 + (2 - 6)^2} \)
\( = \sqrt{36 + 16} = \)
\( 2 \sqrt{13}\)

Here diagonals of ABCD are not equal. … (2)

From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram.