Question

Find the co-ordinates of the point A which divides segment GH internally in the ratio m:n
(i) G $\cong$ (5,8), H $\equiv$($-$7, $-$8), m:n = 4:1
(ii) G $\equiv$ ($-$4, 12), H $\equiv$ (11,$-$3), m:n = 2:3

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Given}:\mathrm{G}\equiv \left(5,8\right)\mathrm{and}\mathrm{H}\equiv \left(-7,-8\right) \\ \mathrm{Here},x_1=5,y_1=8,x_2=-7\mathrm{and}{y}_2=-8. \\ \mathrm{And}m=4\mathrm{and}n=1 \\ \mathrm{Let}\mathrm{A}\equiv \left(x,y\right) \\ \mathrm{By}\mathrm{section}\mathrm{formula}\mathrm{for}\mathrm{internal}\mathrm{division}, \end{gathered}$$

$$\begin{gathered} x=\frac{m{x}_2+n{x}_1}{m+n} \\ \Rightarrow x=\frac{4\times \left(-7\right)+1\times 5}{4+1} \\ \Rightarrow x=\frac{-28+5}{5} \\ \therefore x=\frac{-23}{5} \end{gathered}$$ $$\begin{gathered} y=\frac{m{y}_2+n{y}_1}{m+n} \\ \Rightarrow y=\frac{4\times \left(-8\right)+1\times 8}{4+1} \\ \Rightarrow y=\frac{-32+8}{5} \\ \therefore y=\frac{-24}{5} \end{gathered}$$

$\therefore \text{C}\mathrm{oordinate}\text{s}\mathrm{of}\mathrm{point}\mathrm{A}\equiv \left(\frac{-23}{5},\frac{-24}{5}\right)$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Given}:\mathrm{G}\equiv \left(-4,12\right)\mathrm{and}\mathrm{H}\equiv \left(11,-3\right) \\ \mathrm{Here},x_1=-4,y_1=12,x_2=11\mathrm{and}{y}_2=-3. \\ \mathrm{And}m=2\mathrm{and}n=3 \\ \mathrm{Let}\mathrm{A}\equiv \left(x,y\right) \\ \mathrm{By}\mathrm{section}\mathrm{formula}\mathrm{for}\mathrm{internal}\mathrm{division}, \end{gathered}$$

$$\begin{gathered} x=\frac{m{x}_2+n{x}_1}{m+n} \\ \Rightarrow x=\frac{2\times 11+3\times \left(-4\right)}{2+3} \\ \Rightarrow x=\frac{22-12}{5} \\ \therefore x=2 \end{gathered}$$ $$\begin{gathered} y=\frac{m{y}_2+n{y}_1}{m+n} \\ \Rightarrow y=\frac{2\times \left(-3\right)+3\times 12}{2+3} \\ \Rightarrow y=\frac{-6+36}{5} \\ \therefore y=6 \end{gathered}$$

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