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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.