# Find the Distance of the Point (2, 5) from the Line 3x + Y + 4 = 0 Measured Parallel to a Line Having Slope 3/4. - Mathematics

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

#### Solution

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

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

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

So, the equation of the line passing through (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.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.8 | Q 8 | Page 66