Question

A line a drawn through A (4, −1) parallel to the line 3x − 4y + 1 = 0. Find the coordinates of the two points on this line which are at a distance of 5 units from A.

Answer 1

The slope of the line 3x − 4y + 1 = 0 or $y=\frac{3}{4}x-\frac{1}{4}$ is $\frac{3}{4}$

So, the slope of the required line is also $\frac{3}{4}$ as it is parallel to the given line.

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

Here, $\left(x_1,y_1\right)=A\left(4,-1\right)$

So, the equation of the line passing through A (4, −1) and having slope $\frac{3}{4}$ is

$$\begin{gathered} \frac{x-x_1}{\mathrm{cos\theta }}=\frac{y-y_1}{\mathrm{sin\theta }} \\ \Rightarrow \frac{x-4}{{\displaystyle \frac{4}{5}}}=\frac{y+1}{{\displaystyle \frac{3}{5}}} \\ \Rightarrow 3x-12=4y+4 \\ \Rightarrow 3x-4y-16=0 \end{gathered}$$

Here, AP = r = 5
Thus, the coordinates of P are given by

$$\begin{gathered} x=x_1\pm r\mathrm{cos\theta },y=y_1\pm r\mathrm{sin\theta } \\ \Rightarrow x=4\pm 5\left(\frac{4}{5}\right),y=-1\pm 5\left(\frac{3}{5}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow x=4\pm 4,y=-1\pm 3 \\ \Rightarrow x=8,y=2\mathrm{and}x=0,y=-4 \end{gathered}$$

Hence, the coordinates of the two points at a distance of 5 units from A are (8, 2) and (0, −4).