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 $2x-3y+14=0$ and $5x+5y-7=0$ ?

Answer 1

Given the two lines, $2x-3y+14=0⟶\left(1\right)5x+5y-7=0⟶\left(2\right)$

Solving these lines we have,

$\frac{x}{21-70}=\frac{y}{70+14}=\frac{1}{10+15}\Rightarrow \frac{x}{-49}=\frac{y}{84}=\frac{1}{25}\Rightarrow x=-\frac{49}{25},y=\frac{84}{25}$

So, $(-\frac{49}{25},\frac{84}{25})$ is the point of intersection of the lines $\left(1\right)$ and $\left(2\right)$.

$\therefore$ The equation of the line passing through the origin and the point $(-\frac{49}{25},\frac{84}{25})$ is,

$y-0=\frac{\frac{84}{25}-0}{-\frac{49}{25}-0}(x-0)\Rightarrow y=-\frac{84}{49}x\Rightarrow y=-\frac{12}{7}x\Rightarrow 12x+7y=0$

$\therefore$ Length of the perpendicular from the point $(4,-7)$ to the line $12x+7y=0$ is,

$\left|\frac{12\left(4\right)+7(-7)}{\sqrt{{12}^2+7^2}}\right|=\left|\frac{48-49}{\sqrt{{12}^2+7^2}}\right|=\left|\frac{-1}{\sqrt{193}}\right|=\frac{1}{193}$ units.