Question

Find the distance of the point of intersection of the two straight lines $2x-3y+5=0$ and $3x+4y=0$ from the straight line $5x-2y=0$.

Answer 1

$3x+4y=0x=-\frac{4}{3}y......\left(i\right)2x-3y+5=0$

substituting $x$ from $\left(i\right)$

$2(-\frac{4}{3}y)-3y=-5\frac{8}{3}y+3y=5\frac{17y}{3}=5\Rightarrow y=\frac{15}{17}x=-\frac{4}{3}yx=-\frac{4}{3}.\frac{15}{17}=-\frac{20}{17}$

So the point of intersection is $P(-\frac{20}{17},\frac{15}{17})$

Let length of perpendicular from $P$ to $5x-2y=0$ be $p$

$p=\frac{|5(-\frac{20}{17})-2\left(\frac{15}{17}\right)|}{\sqrt{5^2+{(-2)}^2}}p=\frac{|-\frac{100}{17}-\frac{30}{17}|}{\sqrt{29}}p=\frac{130}{17\sqrt{29}}$