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 \[\left( - \frac{35}{23}, \frac{84}{23} \right)\] .

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

\[y - 0 = \frac{\frac{84}{23} - 0}{\frac{- 35}{23} - 0}\left( x - 0 \right)\]

\[ \Rightarrow y = \frac{84}{- 35}x\]

\[ \Rightarrow y = - \frac{12}{5}x\]

\[ \Rightarrow 12x + 5y = 0\]

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

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