Question

Find the distance of the point $(2,5)$ from the line $3x+y+4=0$ measured parallel to the line $3x-4y+8=0$.

Answer 1

Let $A=(2,5)=(x_1,y_1)$

As we know, it is given that the required line is parallel to $3x-4y+8=0$

$\Rightarrow 4y=3x+8\Rightarrow y=\frac{3}{4}x+8$

$\therefore \tan \theta =\frac{y}{x}=\frac{3}{4}\Rightarrow \sin \theta =\frac{3}{5},\cos \theta =\frac{4}{5}$

so, the equation of line is

$\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }$

$\Rightarrow \frac{x-2}{4/5}=\frac{y-5}{3/5}\Rightarrow 3x-6=4y-20$

$\Rightarrow 3x-4y+14=0$

Let the line $3x-4y+14=0$ cut the line $3x+y+4=0$ at $P$.

Let $AP=x$. Then, the coordinates of $P$ are given by

$\frac{x-2}{4/5}=\frac{y-5}{3/5}=r\Rightarrow x=2+4r/5,y=5+\frac{3r}{5}$

Thus, are coordinates of $P$ are $(2+\frac{4r}{5},5+\frac{3r}{5})$

clearly, $P$ lies on the line $3x+y+4=0$

$\therefore 3(2+\frac{4r}{5})+(5+\frac{3r}{5})+4=0$

$\Rightarrow 6+\frac{12r}{5}+5+\frac{3r}{5}+4=0$

$\Rightarrow 15+\frac{15r}{5}=0$

$\Rightarrow r=-5$

$\therefore AP=|r|=5$

Hence, distance of the point $(2,5)$ from the line $3x+y+4=0$ measured parallel to the line $3x-4y+8=0$ is $5$.