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Question

Find the incentre of the triangle whose vertices are (2,3),(2,5),(4,6).

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Solution

Let
A=(2,3)
B=(2,5)
C=(4,6)
Now by using straight line formula for two points (x1x2)2+(y1y2)2 We get,
BC=(x1x2)2+(y1y2)2
=(2+4)2+(56)2
=22+(11)2
=4+121
=125
=55

Now , again we get
AC=(2+4)2+(63)2
=62+32
=36+9
=45
=35
similarly,
AB=(2+2)2+(3+5)2
=42+82
=16+64
=80
=45
Now for the incentre of the triangle we get
I=(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)

=(55×2+35×(2)+45×(4)55+35+45,55×3+35×(5)+45×655+35+45)
=(1056565125,155155+245125)
=(125125,245125)
=(1,2)
Hence the required points of incentre is =(1,2)


1075122_1053928_ans_d55414ccdcb34343b47ae7d7b305b88a.PNG

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