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.

Solution :

To find the point intersection of the lines x + 2y = 5 and x − 3y = 7, let us solve them.

$$\begin{gathered} \frac{x}{-14-15}=\frac{y}{-5+7}=\frac{1}{-3-2} \\ \Rightarrow x=\frac{29}{5}, y=-\frac{2}{5} \end{gathered}$$

So, the equation of the line passing through $\left(\frac{29}{5},-\frac{2}{5}\right)$ with slope 5 is

$$\begin{gathered} \mathrm{y}+\frac{2}{5}=5\left(\mathrm{x}-\frac{29}{5}\right) \left[\mathrm{as}, y-y_1 = m\left(x-x_1\right)\right] \\ \Rightarrow 5\mathrm{y}+2=25\mathrm{x}-145 \\ \Rightarrow 25\mathrm{x}-5\mathrm{y}-147=0 \end{gathered}$$

$$\begin{gathered} \mathrm{Distance} \mathrm{of} \mathrm{a} \mathrm{point} \left({\mathrm{x}}_1, {\mathrm{y}}_1\right) \mathrm{from} \mathrm{the} \mathrm{line} \mathrm{ax}+\mathrm{by}+\mathrm{c} = 0, \mathrm{is} \\ \mathrm{distance} = \frac{\left|{\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}} \end{gathered}$$

Let d be the perpendicular distance from the point (1, 2) to the line $25x-5y-147=0$


$\therefore d=\left|\frac{25-10-147}{\sqrt{{25}^2+{\displaystyle 5^2}}}\right|=\frac{132}{5\sqrt{26}}$

Hence, the required perpendicular distance is $\frac{132}{5\sqrt{26}}$ units