If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are lengths of the sides $\mathrm{BC}, \mathrm{CA}, \mathrm{AB}$…
- Circumcentre of $\Delta \mathrm{ABC}$
- Incentre of $\Delta \mathrm{ABC}$
- Centroid of $\Delta \mathrm{ABC}$
- Orthocentre of $\Delta \mathrm{ABC}$
Solution
Consider $\mathrm{H}$ to be origin.
Then position vector of the vertices $A, B, C$ are $\bar{a}, \bar{b}, \bar{c}$ respectively.
We have $\mathrm{a} \overline{\mathrm{AH}}+\mathrm{b} \overline{\mathrm{BH}}+\mathrm{c} \overline{\mathrm{CH}}=\overline{0}$ i.e. $\quad \mathrm{a} \overline{\mathrm{a}}+\mathrm{b} \overline{\mathrm{b}}+\mathrm{c} \overline{\mathrm{c}}=0$
i.e. $\frac{a \bar{a}+b \bar{b}+c \bar{c}}{a+b+c}=0$, which is position vector of incentre.
Hence $\mathrm{H}$ is incentre of triangle.Asked in: MHT CET 2020 (13 Oct Shift 1)