Question

(iii) Q $\equiv$ (6, -5) R $\equiv$ (-10, 2) and m:n =3:4

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{Q}\equiv \left(6,-5\right)\mathrm{and}\mathrm{R}\equiv \left(-10,2\right) \\ \mathrm{Here},x_1=6,y_1=-5,x_2=-10\mathrm{and}{y}_2=2. \\ \mathrm{And}m=3\mathrm{and}n=4. \\ \mathrm{Let}\mathrm{P}\equiv \left(x,y\right) \\ \mathrm{By}\mathrm{section}\mathrm{formula}\mathrm{for}\mathrm{internal}\mathrm{division}, \end{gathered}$$

$$\begin{gathered} x=\frac{m{x}_2+n{x}_1}{m+n} \\ \Rightarrow x=\frac{3\times \left(-10\right)+4\times 6}{3+4} \\ \Rightarrow x=\frac{-30+24}{7} \\ \therefore x=\frac{-6}{7} \end{gathered}$$ $$\begin{gathered} y=\frac{m{y}_2+n{y}_1}{m+n} \\ \Rightarrow y=\frac{3\times 2+4\times \left(-5\right)}{3+4} \\ \Rightarrow y=\frac{6-20}{7} \\ \therefore y=-2 \end{gathered}$$

$\therefore \text{C}\mathrm{oordinate}\text{s}\mathrm{of}\mathrm{point}\mathrm{P}\equiv \left(\frac{-6}{7},-2\right)$