In a $∆\mathrm{ABC},$ I is the incentre and $\overline{\mathrm{AI}}$ cuts $\overline{\mathrm{BC}}$ at $\mathrm{D}.$ If $\mathrm{B}=(-1, 2), \mathrm{C}=(9, 7), \mathrm{AB}=14, A\mathrm{C}=21$ then $\mathrm{D}$ is

Given vertices of triangle are

B=(-1, 2), C=(9, 7) and AB=14, AC=21.

We know that in triangle ABC, internal angle bisector of angle A divides the opposite side $\overline{\mathrm{BC}}$ in the ratio AB:AC internally. Where AB:AC=2:3

$$\begin{gathered} \therefore \mathrm{D}=\left[\frac{2\left(9\right)+3(-1)}{2+3}, \frac{2\left(7\right)+3\left(2\right)}{2+3}\right] \\ =\left[\frac{15}{5}, \frac{20}{5}\right] \\ =\left[3, 4\right] \end{gathered}$$