Question

Given 3 points with position vectors $\overline{p_1},\overline{p_2}and\overline{p_3}$ which form the vertices of a triangle with side lengths $a=|\overline{p_2}-\overline{p_1}|,b=|\overline{p_3}-\overline{p_2}|,c=|\overline{p_1}-\overline{p_3}|$. Then the in-centre is given by

• A. $I=\frac{a.\overline{p_1}+a.\overline{p_2}+c.\overline{p_3}}{a+b+c}$
• B. $I=\frac{b.\overline{p_1}+c.\overline{p_2}+a.\overline{p_3}}{a+b+c}$
• C. $I=\frac{c.\overline{p_1}+a.\overline{p_2}+b.\overline{p_3}}{a+b+c}$
• D. $I=\frac{\overline{p_1}+\overline{p_2}+\overline{p_3}}{a+b+c}$

Answer 1

The correct option is B $I=\frac{b.\overline{p_1}+c.\overline{p_2}+a.\overline{p_3}}{a+b+c}$

We know the in-center or inscribed circle centre is given by the weighted average of coordinates by the opposite sides. Here for $\overline{p_1}$ opposite side length will be $|\overline{p_2}-\overline{p_3}|$ which is equal to b. Similarly for $\overline{p_2}$&$\overline{p_3}$ it is c and a.
Therefore Incentre will be the point given by the vector $\frac{b\overline{p_1}+c\overline{p_2}+a\overline{p_3}}{a+b+c}$
$\therefore$ option b is the correct answer.