The incentre of the triangle whose vertices are $\mathrm{P}(0,3,0), \mathrm{Q}(0,0,4)$ and $\mathrm{R}(0,3…

The incentre of the triangle whose vertices are $\mathrm{P}(0,3,0), \mathrm{Q}(0,0,4)$ and $\mathrm{R}(0,3,4)$ is
  1. $\quad(0,3,2)$
  2. $(0,2,3)$
  3. $(2,0,3)$
  4. $(2,3,0)$

Solution

$\begin{array}{ll} & \text { Let } \overline{\mathrm{p}}=3 \hat{\mathrm{j}}, \overline{\mathrm{q}}=4 \hat{\mathrm{k}}, \overline{\mathrm{r}}=3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ \therefore \quad & \overline{\mathrm{PQ}}=-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \overline{\mathrm{QR}}=3 \hat{\mathrm{j}} \\ & \overline{\mathrm{PR}}=4 \hat{\mathrm{k}} \\ & \Rightarrow|\overline{\mathrm{PQ}}|=5,|\overline{\mathrm{QR}}|=3,|\overline{\mathrm{PR}}|=4\end{array}$ Incentre of $\triangle \mathrm{PQR}$ is given by $\begin{aligned} & \frac{|\overline{\mathrm{PQ}}| \overrightarrow{\mathrm{r}}+|\overline{\mathrm{QR}}| \overline{\mathrm{p}}+|\overline{\mathrm{PR}}| \overrightarrow{\mathrm{q}}}{|\overline{\mathrm{PQ}}|+|\overline{\mathrm{QR}}|+|\overline{\mathrm{PR}}|} \\ & =\frac{5(3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+3(3 \hat{\mathrm{j}})+4(4 \hat{\mathrm{k}})}{5+3+4} \\ & =\frac{24 \hat{\mathrm{j}}+36 \hat{\mathrm{k}}}{12} \\ & =2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ \therefore \quad & \text { Incentre } \equiv(0,2,3) \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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