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Question Number 189971 by Rupesh123 last updated on 25/Mar/23

Answered by ajfour last updated on 25/Mar/23

$$y=\frac{3}{2}x$$$$s=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$$$$p=\frac{1}{2}+\frac{s}{\sqrt{2}}\cos \theta =\frac{1}{2}+\frac{\sqrt{13}}{2}\times \frac{1}{\sqrt{2}}\times \frac{3}{\sqrt{13}}$$$$q=\frac{3}{4}-\frac{s}{\sqrt{2}}\sin \theta =\frac{3}{4}-\frac{\sqrt{13}}{2}\times \frac{1}{\sqrt{2}}\times \frac{2}{\sqrt{13}}$$$$\tan \theta =\frac{2}{3}$$$$p=\frac{3\sqrt{2}+2}{4}q=\frac{3-2\sqrt{2}}{4}$$$$x^2={(\frac{3\sqrt{2}+2}{4}-1)}^2+{\left(\frac{3-2\sqrt{2}}{4}\right)}^2$$$$=\frac{18+4-12\sqrt{2}+9+8-12\sqrt{2}}{16}$$$$=\frac{39-24\sqrt{2}}{16}$$$$x=\frac{\sqrt{39-24\sqrt{2}}}{4}$$$$$$

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