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Question Number 183120 by Shrinava last updated on 20/Dec/22

$$\mathrm{I}-\mathrm{Incenter}\mathrm{in}△\mathrm{ABC}$$$$\mathrm{A}(2,2),\mathrm{B}(6,4),\mathrm{C}(4,8),\mathrm{M}(8,6)$$$$\mathrm{Find}:\mathrm{MI}=?$$

Answered by manolex last updated on 21/Dec/22

$$\mathbb{I}=\frac{Aa+\mathit{B}b+Cc}{a+b+c}$$$$a=\parallel \overrightarrow{BC}\parallel =\sqrt{{(4-6)}^2+{(8-4)}^2}=2\sqrt{}5$$$$b=\parallel \overrightarrow{AC}\parallel =\sqrt{2^2+6^2}=2\sqrt{10}$$$$c=\parallel AB\parallel =2\sqrt{5}$$$$I=\frac{(2,2)2\sqrt{5}+(6,4)2\sqrt{10}+(4,8)2\sqrt{5}}{4\sqrt{5}+2\sqrt{10}}$$$$I=\frac{(4\sqrt{5}+12\sqrt{10}+8\sqrt{5};4\sqrt{5}+8\sqrt{10}+16\sqrt{5})}{4\sqrt{5}+2\sqrt{10}}$$$$I=\frac{4\sqrt{5}(3+3\sqrt{2};5+2\sqrt{2})}{2\sqrt{5}(2+\sqrt{2})}$$$$I=2(\frac{3(1+\sqrt{2})}{\sqrt{2}(1+\sqrt{2)}};\frac{5+2\sqrt{2}}{\sqrt{2}(1+\sqrt{2)}})$$$$I=(3\sqrt{2};6-\sqrt{2})$$$$MI=\sqrt{[{(3\sqrt{2}-8)}^2+{(6-\sqrt{2}-6)}^2}$$$$MI=\sqrt{88-48\sqrt{2}}$$$$MI=2\sqrt{2}\sqrt{11-6\sqrt{2}}$$$$MI=2\sqrt{2}(\sqrt{9}-\sqrt{2)}$$$$MI=2\sqrt{18}-4$$$$$$

Answered by manolex last updated on 21/Dec/22

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