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:

$$\frac{x}{21-56}=\frac{y}{70+14}=\frac{1}{8+15}$$

$$\Rightarrow x=-\frac{35}{23},y=\frac{84}{23}$$

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

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

$$y-0=\frac{\frac{84}{23}-0}{\frac{-35}{23}-0}(x-0)$$

$$\Rightarrow y=\frac{84}{-35}x$$

$$\Rightarrow y=-\frac{12}{5}x$$

$$\Rightarrow 12x+5y=0$$

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

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