Answer 1

\[\text { Here }, \left( x_1 , y_1 \right) = A \left( 2, 5 \right), \tan\theta = \frac{3}{4}\]

\[ \Rightarrow sin\theta = \frac{3}{\sqrt{3^2 + 4^2}} \text { and } cos\theta = \frac{4}{\sqrt{3^2 + 4^2}}\]

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

So, the equation of the line passing through A (2, 5) and having slope  \[\frac{3}{4}\] 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 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 \[\frac{x - 2}{\frac{4}{5}} = \frac{y - 5}{\frac{3}{5}} = r\] 

\[\Rightarrow x = 2 + \frac{4r}{5} \text { and }y = 5 + \frac{3r}{5}\]

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.

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

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

\[ \Rightarrow 3r = - 15\]

\[ \Rightarrow r = - 5\]

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