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the coordinate of the incenter of the triangle whose vertices are (4,-2 ) (-2,4) and (5,5)

sach , 11 Years ago
Grade 12
anser 1 Answers
Latika Leekha

Last Activity: 10 Years ago

The given coordinates are (4, -2), (-2,4) and (5,5).
Let us denote them as A(4, -2), B(-2,4) and C(5,5).
Then a = |BC| = \sqrt{}(5+2)2 + (5-4)2 = \sqrt{}(72+12) = \sqrt{}50 = 5\sqrt{}2
b = |CA| = \sqrt{}(4-5)2 + (-2-5)2 = \sqrt{}(72+12) = \sqrt{}50 = 5\sqrt{}2
c = |AB| = \sqrt{}(-2-4)2 + (4+2)2 = \sqrt{}(36 + 36) = \sqrt{}72 = 6\sqrt{}2
Coordinates of incentre of traingle ABC are
[(ax1 + bx2 + cx3) / (a+b+c) , (ay1 + by2 + cy3) / (a+b+c)]
i.e. [ {5\sqrt{}2 .4 + 5\sqrt{}2 . (-2) + 6\sqrt{}2 .5 }/ 16\sqrt{}2 , {5\sqrt{}2 . (-2) + 5\sqrt{}2 . 4 + 6\sqrt{}2 .5 }/ 16\sqrt{}2 ],
Hence, this gives ( 40\sqrt{}2/ 16\sqrt{}2 , 40\sqrt{}2/ 16\sqrt{}2)
Hence, the coordinates of the incentre are (5/2, 5/2).
Thanks & Regards
Latika Leekha
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