Question

In what ratio does the point (1,3) divide the line segment joining the points (3, 6) and (-5, -6) internally.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{P}\equiv \left(1,3\right)\mathrm{divide}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{AB}\mathrm{in}\mathrm{the}\mathrm{ratio}\mathrm{m}:\mathrm{n}. \\ \text{H}\mathrm{ere},\mathrm{A}\equiv \left(3,6\right)\mathrm{and}\mathrm{B}\equiv \left(-5,-6\right). \\ \mathrm{Here},x_1=3,y_1=6,x_2=-5\mathrm{and}{y}_2=-6. \\ x=1\mathrm{and}y=3 \\ \mathrm{By}\mathrm{section}\mathrm{formula}\mathrm{for}\mathrm{internal}\mathrm{division}, \end{gathered}$$

$$\begin{gathered} x=\frac{m{x}_2+n{x}_1}{m+n} \\ \Rightarrow 1=\frac{m\times \left(-5\right)+n\times 3}{m+n} \\ \Rightarrow m+n=-5m+3n \\ \Rightarrow 6m=2n \\ \Rightarrow 3m=n \\ \Rightarrow \frac{m}{n}=\frac{1}{3} \\ \therefore m:n=1:3 \end{gathered}$$ $$\begin{gathered} Andy=\frac{m{y}_2+n{y}_1}{m+n} \\ \Rightarrow 3=\frac{m\times \left(-6\right)+n\times 6}{m+n} \\ \Rightarrow 3m+3n=-6m+6n \\ \Rightarrow 9m=3n \\ \Rightarrow 3m=n \\ \Rightarrow \frac{m}{n}=\frac{1}{3} \\ \therefore m:n=1:3 \end{gathered}$$

$$\begin{gathered} \mathrm{In}\mathrm{both}\mathrm{the}\mathrm{case}\text{s},m:n=1:3. \\ \therefore \text{R}\mathrm{equired}\mathrm{ratio}=1:3. \end{gathered}$$