Question

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

(i) A(0, −4), B(6, 2) C(3, 5) and D(−3, −1)
(ii) A(2, −2), B(14, 10) C(11, 13) and D(−1, 1)
(iii) A(−4, −1), B(−2, −4) C(−3, 2) and D(0, −1)

Answer 1

(i) The given points are A(0, −4), B(6, 2) C(3, 5) and D(−3, −1).
$$\begin{gathered} AB=\sqrt{{\left(6-0\right)}^2+{\left\{2-\left(-4\right)\right\}}^2}=\sqrt{{\left(6\right)}^2+{\left(6\right)}^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}\mathrm{units} \\ BC=\sqrt{{\left(3-6\right)}^2+{\left(5-2\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ CD=\sqrt{{\left(-3-3\right)}^2+{\left(-1-5\right)}^2}=\sqrt{{\left(-6\right)}^2+{\left(-6\right)}^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}\mathrm{units} \\ AD=\sqrt{{\left(-3-0\right)}^2+{\left\{-1-\left(-4\right)\right\}}^2}=\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ \mathrm{Thus},AB=CD=\sqrt{10}\mathrm{units}\mathrm{and}BC=AD=\sqrt{5}\mathrm{units} \\ \mathrm{Also},AC=\sqrt{{\left(3-0\right)}^2+{\left\{5-\left(-4\right)\right\}}^2}=\sqrt{{\left(3\right)}^2+{\left(9\right)}^2}=\sqrt{9+81}=\sqrt{90}=3\sqrt{10}\mathrm{units} \\ BD=\sqrt{{\left(-3-6\right)}^2+{\left(-1-2\right)}^2}=\sqrt{{\left(-9\right)}^2+{\left(-3\right)}^2}=\sqrt{81+9}=\sqrt{90}=3\sqrt{10}\mathrm{units} \end{gathered}$$
Also, diagonal AC = diagonal BD.
Hence, the given points form a rectangle.

(ii) The given points are A(2, −2), B(14, 10) C(11, 13) and D(−1, 1).
$$\begin{gathered} AB=\sqrt{{\left(14-2\right)}^2+{\left\{10-\left(-2\right)\right\}}^2}=\sqrt{{\left(12\right)}^2+{\left(12\right)}^2}=\sqrt{144+144}=\sqrt{288}=12\sqrt{2}\mathrm{units} \\ BC=\sqrt{{\left(11-14\right)}^2+{\left(13-10\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ CD=\sqrt{{\left(-1-11\right)}^2+{\left(1-13\right)}^2}=\sqrt{{\left(-12\right)}^2+{\left(-12\right)}^2}=\sqrt{144+144}=\sqrt{288}=12\sqrt{2}\mathrm{units} \\ AD=\sqrt{{\left(-1-2\right)}^2+{\left\{1-\left(-2\right)\right\}}^2}=\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\mathrm{units} \\ \mathrm{Thus},AB=CD=12\sqrt{2}\mathrm{units}\mathrm{and}BC=AD=3\sqrt{2}\mathrm{units} \\ \mathrm{Also},AC=\sqrt{{\left(11-2\right)}^2+{\left\{13-\left(-2\right)\right\}}^2}=\sqrt{{\left(9\right)}^2+{\left(15\right)}^2}=\sqrt{81+225}=\sqrt{306}=3\sqrt{34}\mathrm{units} \\ BD=\sqrt{{\left(-1-14\right)}^2+{\left(1-10\right)}^2}=\sqrt{{\left(-15\right)}^2+{\left(-9\right)}^2}=\sqrt{81+225}=\sqrt{306}=3\sqrt{34}\mathrm{units} \end{gathered}$$
Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.

(iii) The given points are A(−4, −1), B(−2, −4) C(4, 0) and D(2, 3).
$$\begin{gathered} AB=\sqrt{{\left\{-2-\left(-4\right)\right\}}^2+{\left\{-4-\left(-1\right)\right\}}^2}=\sqrt{{\left(2\right)}^2+{\left(-3\right)}^2}=\sqrt{4+9}=\sqrt{13}\mathrm{units} \\ BC=\sqrt{{\left\{4-\left(-2\right)\right\}}^2+{\left\{0-\left(-4\right)\right\}}^2}=\sqrt{{\left(6\right)}^2+{\left(4\right)}^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\mathrm{units} \\ CD=\sqrt{{\left(2-4\right)}^2+{\left(3-0\right)}^2}=\sqrt{{\left(-2\right)}^2+{\left(3\right)}^2}=\sqrt{4+9}=\sqrt{13}\mathrm{units} \\ AD=\sqrt{{\left\{2-\left(-4\right)\right\}}^2+{\left\{3-\left(-1\right)\right\}}^2}=\sqrt{{\left(6\right)}^2+{\left(4\right)}^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\mathrm{units} \\ \mathrm{Thus},AB=CD=\sqrt{13}\mathrm{units}\mathrm{and}BC=AD=2\sqrt{13}\mathrm{units} \\ \mathrm{Also},AC=\sqrt{{\left\{4-\left(-4\right)\right\}}^2+{\left\{0-\left(-1\right)\right\}}^2}=\sqrt{{\left(8\right)}^2+{\left(1\right)}^2}=\sqrt{64+1}=\sqrt{65}\mathrm{units} \\ BD=\sqrt{{\left\{2-\left(-2\right)\right\}}^2+{\left\{3-\left(-4\right)\right\}}^2}=\sqrt{{\left(4\right)}^2+{\left(7\right)}^2}=\sqrt{16+49}=\sqrt{65}\mathrm{units} \end{gathered}$$
Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.