I-Incenter-in-ABC-A-2-2-B-6-4-C-4-8-M-8-6-Find-MI-

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

I - Incenter in △ ABC

A (2, 2), B (6, 4), C (4, 8), M (8, 6)

Find : MI = ?

Answered by manolex last updated on 21/Dec/22

I = \frac{A a + B b + C c}{a + b + c}

a = ∥ \vec{B C} ∥= \sqrt{{(4 - 6)}^{2} + {(8 - 4)}^{2}} = 2 \sqrt{} 5

b =∥ \vec{A C} ∥= \sqrt{2^{2} + 6^{2}} = 2 \sqrt{10}

c =∥ A B ∥= 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})

M I = \sqrt{[{(3 \sqrt{2} - 8)}^{2} + {(6 - \sqrt{2} - 6)}^{2}}

M I = \sqrt{88 - 48 \sqrt{2}}

M I = 2 \sqrt{2} \sqrt{11 - 6 \sqrt{2}}

M I = 2 \sqrt{2} (\sqrt{9} - \sqrt{2)}

M I = 2 \sqrt{18} - 4

Answered by manolex last updated on 21/Dec/22

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