Draw the quadrilateral whose vertices are (i) (1, 1), (2, 4), (8, 4) and (10, 1) (ii) (-2, -2), (-4, 2), (-6, -2) and (-4, -6). Name the type of quadrilateral so formed in each case.

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Please find below the solution to the asked query:



$$\begin{gathered} \mathrm{Slope} \mathrm{of} \mathrm{AB}=\frac{1-1}{10-1}=0 \\ \mathrm{Slope} \mathrm{of} \mathrm{CD}=\frac{4-4}{8-2}=0 \\ \mathrm{As} \mathrm{slope} \mathrm{of} \mathrm{AB}=\mathrm{Slope} \mathrm{of} \mathrm{CD} \\ \mathrm{So} \mathrm{AB}\parallel \mathrm{CD},\mathrm{hence} \\ \mathrm{ABCD} \mathrm{is} \mathrm{a} \mathrm{trapezium}. \end{gathered}$$



$$\begin{gathered} \mathrm{Using} \mathrm{distance} \mathrm{formula} \mathrm{we} \mathrm{have} \\ \mathrm{AB}=\sqrt{{\left(-2-2\right)}^2+{\left\{-2-\left(-4\right)\right\}}^2} \\ =\sqrt{16+4} \\ =2\sqrt{5} \\ \mathrm{BC}=\sqrt{{\left\{-4-\left(-2\right)\right\}}^2+{\left\{-6-\left(-2\right)\right\}}^2} \\ =\sqrt{4+16} \\ =2\sqrt{5} \\ \mathrm{AD}=\sqrt{{\left\{-4-\left(-6\right)\right\}}^2+{\left\{2-\left(-2\right)\right\}}^2} \\ =\sqrt{4+16} \\ =2\sqrt{5} \\ \mathrm{CD}=\sqrt{{\left\{-6-\left(-4\right)\right\}}^2+{\left\{-2-\left(-6\right)\right\}}^2} \\ =\sqrt{4+16} \\ =2\sqrt{5} \\ \mathrm{As} \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA} \\ \mathrm{Hence} \mathrm{ABCD}\hspace{0.17em}\mathrm{is} \mathrm{a} \mathrm{rhombus}. \end{gathered}$$
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