Question

Show that the following points are the vertices of a square.

(i) A(3, 2), B(0, 5), C(−3, 2) and D(0, −1)
(ii) A(6, 2), B(2, 1), C(1, 5) and D(5, 6)
(iii) P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1)

Answer 1

(i) The given points are A(3, 2), B(0, 5), C(−3, 2) and D(0, −1).
$$\begin{gathered} AB=\sqrt{{\left(0-3\right)}^2+{\left(5-2\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ BC=\sqrt{{\left(-3-0\right)}^2+{\left(2-5\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(-3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ CD=\sqrt{{\left(0+3\right)}^2+{\left(-1-2\right)}^2}=\sqrt{{\left(3\right)}^2+{\left(-3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ DA=\sqrt{{\left(0-3\right)}^2+{\left(-1-2\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(-3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ \mathrm{Therefore},AB=BC=CD=DA=3\sqrt{2}\mathrm{units} \\ \mathrm{Also},AC=\sqrt{{\left(-3-3\right)}^2+{\left(2-2\right)}^2}=\sqrt{{\left(-6\right)}^2+{\left(0\right)}^2}=\sqrt{36}=6\mathrm{units} \\ BD=\sqrt{{\left(0-0\right)}^2+{\left(-1-5\right)}^2}=\sqrt{{\left(0\right)}^2+{\left(-6\right)}^2}=\sqrt{36}=6\mathrm{units} \\ \mathrm{Thus},\mathrm{diagonal}AC=\mathrm{diagonal}BD \end{gathered}$$
Therefore, the given points form a square.


(ii) The given points are A(6, 2), B(2, 1), C(1, 5) and D(5, 6).
$$\begin{gathered} AB=\sqrt{{\left(2-6\right)}^2+{\left(1-2\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(-1\right)}^2}=\sqrt{16+1}=\sqrt{17}\mathrm{units} \\ BC=\sqrt{{\left(1-2\right)}^2+{\left(5-1\right)}^2}=\sqrt{{\left(-1\right)}^2+{\left(-4\right)}^2}=\sqrt{1+16}=\sqrt{17}\mathrm{units} \\ CD=\sqrt{{\left(5-1\right)}^2+{\left(6-5\right)}^2}=\sqrt{{\left(4\right)}^2+{\left(1\right)}^2}=\sqrt{16+1}=\sqrt{17}\mathrm{units} \\ DA=\sqrt{{\left(5-6\right)}^2+{\left(6-2\right)}^2}=\sqrt{{\left(1\right)}^2+{\left(4\right)}^2}=\sqrt{1+16}=\sqrt{17}\mathrm{units} \\ \mathrm{Therefore}AB=BC=CD=DA=\sqrt{17}\mathrm{units} \\ \mathrm{Also}AC=\sqrt{{\left(1-6\right)}^2+{\left(5-2\right)}^2}=\sqrt{{\left(-5\right)}^2+{\left(3\right)}^2}=\sqrt{25+9}=\sqrt{34}\mathrm{units} \\ BD=\sqrt{{\left(5-2\right)}^2+{\left(6-1\right)}^2}=\sqrt{{\left(3\right)}^2+{\left(5\right)}^2}=\sqrt{9+25}=\sqrt{34}\mathrm{units} \\ \mathrm{Thus},\mathrm{diagonal}AC=\mathrm{diagonal}BD \end{gathered}$$
Therefore, the given points form a square.


(iii) The given points are P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1).
$$\begin{gathered} PQ=\sqrt{{\left(3-0\right)}^2+{\left(1+2\right)}^2}=\sqrt{{\left(3\right)}^2+{\left(3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ QR=\sqrt{{\left(0-3\right)}^2+{\left(4-1\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ RS=\sqrt{{\left(-3-0\right)}^2+{\left(1-4\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(-3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ SP=\sqrt{{\left(-3-0\right)}^2+{\left(1+2\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ \mathrm{Therefore},PQ=QR=RS=SP=3\sqrt{2}\mathrm{units} \\ \mathrm{Also},PR=\sqrt{{\left(0-0\right)}^2+{\left(4+2\right)}^2}=\sqrt{{\left(0\right)}^2+{\left(6\right)}^2}=\sqrt{36}=6\mathrm{units} \\ QS\hspace{0.17em}=\sqrt{{\left(-3-3\right)}^2+{\left(1-1\right)}^2}=\sqrt{{\left(-6\right)}^2+{\left(0\right)}^2}=\sqrt{36}=6\mathrm{units} \\ \mathrm{Thus},\mathrm{diagonal}PR=\mathrm{diagonal}QS \end{gathered}$$
Therefore, the given points form a square.