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

\[\Rightarrow 4y = 3x + 8\]

\[ \Rightarrow y = \frac{3}{4}x + 2\]

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

So, the equation of the line is

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

\[ \Rightarrow \frac{x - 2}{\frac{4}{5}} = \frac{y - 5}{\frac{3}{5}}\]

\[ \Rightarrow 3x - 6 = 4y - 20\]

\[ \Rightarrow 3x - 4y + 14 = 0\]

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

Let AP = r
Then, the coordinates of P are given by \[\frac{x - 2}{\frac{4}{5}} = \frac{y - 5}{\frac{3}{5}} = r\] \[\Rightarrow x = 2 + \frac{4r}{5}, y = 5 + \frac{3r}{5}\]

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.

\[\therefore 3\left( 2 + \frac{4r}{5} \right) + 5 + \frac{3r}{5} + 4 = 0\]

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

\[ \Rightarrow 15 + \frac{15r}{5} = 0\]

\[ \Rightarrow r = - 5\]

∴ AP = \[\left| r \right|\] = 5