Question

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

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

$$\begin{gathered} y+\frac{2}{5}=5\left(x-\frac{29}{5}\right) \\ \Rightarrow 5y+2=25x-145 \\ \Rightarrow 25x-5y-147=0 \end{gathered}$$

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 $\frac{132}{5\sqrt{26}}$