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

Here, $\left(x_1,y_1\right)=A\left(2,5\right)$
It is given that the required line is parallel to 3x −4y + 8 = 0

$$\begin{gathered} \Rightarrow 4y=3x+8 \\ \Rightarrow y=\frac{3}{4}x+2 \end{gathered}$$

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

So, the equation of the line is

$$\begin{gathered} \frac{x-x_1}{\mathrm{cos\theta }}=\frac{y-y_1}{\mathrm{sin\theta }} \\ \Rightarrow \frac{x-2}{{\displaystyle \frac{4}{5}}}=\frac{y-5}{{\displaystyle \frac{3}{5}}} \\ \Rightarrow 3x-6=4y-20 \\ \Rightarrow 3x-4y+14=0 \end{gathered}$$

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

Let AP = r
Then, the coordinates of P are given by

$\frac{x-2}{{\displaystyle \frac{4}{5}}}=\frac{y-5}{{\displaystyle \frac{3}{5}}}=r$

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

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

Clearly, P lies on the line 3x + y + 4 = 0.

$$\begin{gathered} \therefore 3\left(2+\frac{4r}{5}\right)+5+\frac{3r}{5}+4=0 \\ \Rightarrow 6+\frac{12r}{5}+5+\frac{3r}{5}+4=0 \\ \Rightarrow 15+\frac{15r}{5}=0 \\ \Rightarrow r=-5 \end{gathered}$$

∴ AP = $\left|r\right|$ = 5