a and b are the points (-2,0) and (0,5) . find the coordinates of two points c and d such that abcd is a square?

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{A}\left(-2,0\right) \mathrm{and} \mathrm{B}\left(0,5\right) \\ \mathrm{Equation} \mathrm{of} \mathrm{AB} \mathrm{is}: \\ \frac{\mathrm{x}}{\left(-2\right)}+\frac{\mathrm{y}}{5}=1 \\ \Rightarrow 5\mathrm{x}-2\mathrm{y}=-10 \\ \Rightarrow 5\mathrm{x}-2\mathrm{y}+10=0 ;\left(\mathrm{i}\right) \\ \mathrm{Slope}=\mathrm{tan\alpha }=-\frac{\mathrm{Coefficient} \mathrm{of} \mathrm{x}}{\mathrm{Coefficient} \mathrm{of} \mathrm{y}}=-\frac{\left(5\right)}{\left(-2\right)}=\frac{5}{2} \\ \Rightarrow \mathrm{cos\alpha }=\frac{2}{\sqrt{29}} \mathrm{and} \mathrm{sin\alpha }=\frac{5}{\sqrt{29}} \\ \mathrm{Side} \mathrm{of} \mathrm{square}=\mathrm{AB}=\sqrt{{\left(-2-0\right)}^2+{\left(0-5\right)}^2}=\sqrt{29} \\ \mathrm{Slope} \mathrm{of} \mathrm{line} \mathrm{perpendicular} \mathrm{to} \left(\mathrm{i}\right) \\ \mathrm{m}=-\frac{1}{\mathrm{tan\alpha }}=-\frac{2}{5}=\mathrm{tan\theta }=\frac{\mathrm{Perpendicular} }{\mathrm{Base}} \\ \mathrm{hypotenuse}=\sqrt{5^2+2^2}=\sqrt{29} \\ \Rightarrow \mathrm{cos\theta }=-\frac{5}{\sqrt{29}} \mathrm{and} \mathrm{sin\theta }=\frac{2}{\sqrt{29}} \left[\mathrm{sine} \mathrm{is} \mathrm{positive} \mathrm{in} \mathrm{second} \mathrm{quadrant}\right] \\ \mathrm{From} \mathrm{point} \mathrm{A}\left(-2,0\right)=\left(\mathrm{h},\mathrm{k}\right)\hspace{0.17em}\mathrm{we} \mathrm{have} \mathrm{to} \mathrm{move} \mathrm{r}=\sqrt{29} \mathrm{units} \mathrm{upwards} \mathrm{along} \\ \mathrm{slope} \mathrm{m}=-\frac{2}{5}=\mathrm{tan\theta } \\ \mathrm{Co}-\mathrm{ordinates} \mathrm{of} \mathrm{D} \mathrm{are}: \\ \left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{h}+\mathrm{rcos\theta },\mathrm{k}+\mathrm{rsin\theta }\right) \\ =\left(-2+\sqrt{29}\left(-\frac{5}{\sqrt{29}}\right),0+\sqrt{29}\left(\frac{2}{\sqrt{29}}\right)\right) \\ =\left(-2-5,0+2\right) \\ =\left(-7,2\right) \\ \mathrm{Similarly} \\ \mathrm{From} \mathrm{point} \mathrm{A}\left(0,5\right)=\left(\mathrm{h},\mathrm{k}\right)\hspace{0.17em}\mathrm{we} \mathrm{have} \mathrm{to} \mathrm{move} \mathrm{r}=\sqrt{29} \mathrm{units} \mathrm{upwards} \mathrm{along} \\ \mathrm{slope} \mathrm{m}=-\frac{2}{5}=\mathrm{tan\theta } \\ \mathrm{Co}-\mathrm{ordinates} \mathrm{of} \mathrm{C} \mathrm{are}: \\ \left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{h}+\mathrm{rcos\theta },\mathrm{k}+\mathrm{rsin\theta }\right) \\ =\left(0+\sqrt{29}\left(-\frac{5}{\sqrt{29}}\right),5+\sqrt{29}\left(\frac{2}{\sqrt{29}}\right)\right) \\ =\left(0-5,5+2\right) \\ =\left(-5,7\right) \\ \mathbf{Hence}\mathbf{ }\mathbf{coordinates}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{ }\mathbf{are}\mathbf{ }\mathbf{C}\left(\mathbf{-}\mathbf{5}\mathbf{,}\mathbf{7}\right)\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{D}\mathbf{ }\mathbf{are}\mathbf{ }\mathbf{D}\left(\mathbf{-}\mathbf{7}\mathbf{,}\mathbf{2}\right) \\ \mathrm{You} \mathrm{can} \mathrm{find} \mathrm{one} \mathrm{more} \mathrm{value} \mathrm{for} \mathrm{C}\hspace{0.17em}\mathrm{and} \mathrm{D} \mathrm{by} \mathrm{using} \left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{h}-\mathrm{rcos\theta },\mathrm{k}-\mathrm{rsin\theta }\right). \end{gathered}$$

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