Question

Find the distance of the point (2, 5) from the line 3x + y + 4 = 0 measured parallel to a line having slope 3/4.

Answer 1

$$\begin{gathered} \mathrm{Here},\left(x_1,y_1\right)=A\left(2,5\right),\tan \mathrm{\theta }=\frac{3}{4} \\ \Rightarrow \mathrm{sin\theta }=\frac{3}{\sqrt{3^2+4^2}}\mathrm{and}\mathrm{cos\theta }=\frac{4}{\sqrt{3^2+4^2}} \\ \Rightarrow \mathrm{sin\theta }=\frac{3}{5}\mathrm{and}\mathrm{cos\theta }=\frac{4}{5} \end{gathered}$$

So, the equation of the line passing through A (2, 5) and having slope $\frac{3}{4}$ 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 3x − 4y + 14 = 0 intersect the line 3x + y + 4 = 0 at point P.
Let AP = r
Then, the coordinates of P are given by

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

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

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.

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

Hence, the distance of the point (2, 5) from the line 3x + y + 4 = 0 is 5.