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Question Number 183120 by Shrinava last updated on 20/Dec/22
I−Incenterin△ABCA(2,2),B(6,4),C(4,8),M(8,6)Find:MI=?
Answered by manolex last updated on 21/Dec/22
I=Aa+Bb+Cca+b+ca=∥BC→∥=(4−6)2+(8−4)2=25b=∥AC→∥=22+62=210c=∥AB∥=25I=(2,2)25+(6,4)210+(4,8)2545+210I=(45+1210+85;45+810+165)45+210I=45(3+32;5+22)25(2+2)I=2(3(1+2)2(1+2);5+222(1+2))I=(32;6−2)MI=[(32−8)2+(6−2−6)2MI=88−482MI=2211−62MI=22(9−2)MI=218−4
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