Question

Find the length of the perpendicular from the point (4, −7) to the line joining the origin and the point of intersection of the lines 2x − 3y + 14 = 0 and 5x + 4y − 7 = 0.

Answer 1

Solving the lines 2x − 3y + 14 = 0 and 5x + 4y − 7 = 0 we get:

$$\begin{gathered} \frac{x}{21-56}=\frac{y}{70+14}=\frac{1}{8+15} \\ \Rightarrow x=-\frac{35}{23},y=\frac{84}{23} \end{gathered}$$

So, the point of intersection of 2x − 3y + 14 = 0 and 5x + 4y − 7 = 0 is $\left(-\frac{35}{23},\frac{84}{23}\right)$.

The equation of the line passing through the origin and the point $\left(-\frac{35}{23},\frac{84}{23}\right)$ is

$$\begin{gathered} y-0=\frac{{\displaystyle \frac{84}{23}}-0}{{\displaystyle \frac{-35}{23}}-0}\left(x-0\right) \\ \Rightarrow y=\frac{84}{-35}x \\ \Rightarrow y=-\frac{12}{5}x \\ \Rightarrow 12x+5y=0 \end{gathered}$$

Let d be the perpendicular distance of the line 12x + 5y = 0 from the point (4, −7)

$\therefore d=\left|\frac{48-35}{\sqrt{{12}^2+5^2}}\right|=\frac{13}{13}=1$