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. 

\[\therefore \tan\theta = \frac{3}{4} \Rightarrow sin\theta = \frac{3}{5} \text { and } cos\theta = \frac{4}{5}\]

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  

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

\[ \Rightarrow \frac{x - 4}{\frac{4}{5}} = \frac{y + 1}{\frac{3}{5}}\]

\[ \Rightarrow 3x - 12 = 4y + 4\]

\[ \Rightarrow 3x - 4y - 16 = 0\]

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

\[x = x_1 \pm r\text { cos } \theta, y = y_1 \pm r\text { sin }\theta\]

\[ \Rightarrow x = 4 \pm 5\left( \frac{4}{5} \right), y = - 1 \pm 5\left( \frac{3}{5} \right)\]

\[\Rightarrow x = 4 \pm 4, y = - 1 \pm 3\]

\[ \Rightarrow x = 8, y = 2 \text { and } x = 0, y = - 4\]

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