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Question

Given 3 points with position vectors ¯p1,¯p2 and ¯p3 which form the vertices of a triangle with side lengths a=|¯p2¯p1|,b=|¯p3¯p2|,c=|¯p1¯p3|. Then the in-centre is given by

A
I=a.¯p1+a.¯p2+c.¯p3a+b+c
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B
I=b.¯p1+c.¯p2+a.¯p3a+b+c
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C
I=c.¯p1+a.¯p2+b.¯p3a+b+c
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D
I=¯p1+¯p2+¯p3a+b+c
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Solution

The correct option is B I=b.¯p1+c.¯p2+a.¯p3a+b+c
We know the in-center or inscribed circle centre is given by the weighted average of coordinates by the opposite sides. Here for ¯p1 opposite side length will be |¯p2¯p3| which is equal to b. Similarly for ¯p2&¯p3 it is c and a.
Therefore Incentre will be the point given by the vector b¯p1+c¯p2+a¯p3a+b+c
option b is the correct answer.

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