Question

7. Find the distance of the point (1,2) from the straight line with slope 5 and passing through the point of intersection of x+2y=5 and x-3y=7

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{lines}\mathrm{are}: \\ \mathrm{x}+2\mathrm{y}=5.....\left(1\right) \\ \mathrm{x}-3\mathrm{y}=7.....\left(2\right) \\ \mathrm{Now},\mathrm{subtracting}\left(2\right)\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get} \\ 5\mathrm{y}=-2\Rightarrow \mathrm{y}=-\frac{2}{5} \\ \mathrm{Put}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{x}+2\times \left(-\frac{2}{5}\right)=5 \\ \Rightarrow \mathrm{x}-\frac{4}{5}=5 \\ \Rightarrow \mathrm{x}=5+\frac{4}{5} \\ \Rightarrow \mathrm{x}=\frac{29}{5} \\ \mathrm{So},\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{is}\left(\frac{29}{5},-\frac{2}{5}\right). \\ \mathrm{Now},\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{straight}\mathrm{line}\mathrm{passing}\mathrm{through}\left(\frac{29}{5},-\frac{2}{5}\right)\mathrm{and}\mathrm{having}\mathrm{slope}5\mathrm{is}, \\ \mathrm{y}+\frac{2}{5}=5\left(\mathrm{x}-\frac{29}{5}\right) \\ \Rightarrow \frac{5\mathrm{y}+2}{5}=5\left[\frac{5\mathrm{x}-29}{5}\right] \\ \Rightarrow 5\mathrm{y}+2=25\mathrm{x}-145 \\ \Rightarrow 25\mathrm{x}-5\mathrm{y}=147.....\left(3\right) \\ \mathrm{Now},\mathrm{distance}\mathrm{of}\left(1,2\right)\mathrm{from}\left(3\right)\mathrm{is}, \\ \mathrm{distance}=\frac{\left|25\times 1-5\times 2-147\right|}{\sqrt{{25}^2+{\left(-5\right)}^2}} \\ =\frac{132}{\sqrt{625+25}} \\ =\frac{132}{\sqrt{650}} \\ =\frac{132}{5\sqrt{26}} \end{gathered}$$