Question.

(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)

Solution:

(i) A(–1, –2), B(1, 0), C(–1, 2), D(–3, 0)

Determine distances : AB, BC, CD, DA, AC and BD.

$\mathrm{AB}=\sqrt{(1+1)^{2}+(0+2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

$\mathrm{BC}=\sqrt{(-1-1)^{2}+(2-0)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

$C D=\sqrt{(-3+1)^{2}+(0-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

$D A=\sqrt{(-1+3)^{2}+(-2-0)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

AB = BC = CD = DA

The sides of the quadrilateral are equal ............(1)

$\left.\begin{array}{l}\mathrm{AC}=\sqrt{(\mathbf{1}+\mathbf{1})^{2}+(\mathbf{2}+\mathbf{2})^{2}}=\sqrt{\mathbf{0}+\mathbf{1 6}}=4 \\ \mathrm{BD}=\sqrt{(\mathbf{3}-\mathbf{1})^{2}+(\mathbf{0}-\mathbf{0})^{2}}=\sqrt{\mathbf{1 6}+\mathbf{0}}=4\end{array}\right\}$

Diagonal AC = Diagonal BD..........(2)

From (1) and (2) we conclude that ABCD is a square.

(ii) Let the points be $\mathrm{A}(-3,5), \mathrm{B}(3,1), \mathrm{C}(0,3)$ and $\mathrm{D}(-1,-4)$.

$\therefore A B=\sqrt{(3-(-3))^{2}+(1-5)^{2}}$

$=\sqrt{6^{2}+(-4)^{2}}=\sqrt{36+16}$

$=\sqrt{52}=2 \sqrt{13}$

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

$C D=\sqrt{(-1-0)^{2}+(-4-3)^{2}}=\sqrt{(-1)^{2}+(-7)^{2}}$

$=\sqrt{1+49}=\sqrt{50}$

$D A=\sqrt{|-3-(-1)|^{2}+[5-(-4)]^{2}}$

$=\sqrt{(-2)^{2}+(9)^{2}}$

$=\sqrt{4+81}=\sqrt{85}$

$A C=\sqrt{(0-(-3))^{2}+(3-5)^{2}}=\sqrt{(3)^{2}+(-2)^{2}}$

$=\sqrt{9+4}=\sqrt{13}$

$B D=\sqrt{(-1-3)^{2}+(-4-1)^{2}}=\sqrt{(-4)^{2}+(-5)^{2}}$

$=\sqrt{16+25}=\sqrt{41}$

We see that $\sqrt{\mathbf{1 3}}+\sqrt{\mathbf{1 3}}=\mathbf{2} \sqrt{\mathbf{1 3}}$

i.e., AC + BC = AB

A, B and C are collinear. Thus, ABCD is not a quadrilateral.

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

$\therefore A B=\sqrt{(7-4)^{2}+(6-5)^{2}}=\sqrt{3^{2}+1^{2}}=\sqrt{10}$

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

$=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{18}$

$C D=\sqrt{(\mathbf{1}-\mathbf{4})^{2}+(\mathbf{2}-\mathbf{3})^{2}}$

$=\sqrt{(-3)^{2}+(-1)^{2}}=\sqrt{10}$

DA $=\sqrt{(\mathbf{1}-\mathbf{4})^{2}+(\mathbf{2}-\mathbf{5})^{2}}=\sqrt{\mathbf{9}+\mathbf{9}}=\sqrt{\mathbf{1 8}}$

$A C=\sqrt{(4-4)^{2}+(3-5)^{2}}=\sqrt{0+(-2)^{2}}=2$

$\mathrm{BD}=\sqrt{(1-7)^{2}+(2-6)^{2}}=\sqrt{36+16}=\sqrt{52}$

Since, AB = CD, BC = DA [opposite sides of the quadrilateral are equal]

And $\mathrm{AC} \neq \mathrm{BD} \Rightarrow$ Diagonals are unequal.

$\therefore$ ABCD is a parallelogram.