# Find the Distance of the Point (2, 5) from the Line 3x + Y + 4 = 0 Measured Parallel to the Line 3x − 4y + 8 = 0. - Mathematics

Find the distance of the point (2, 5) from the line 3x + y + 4 = 0 measured parallel to the line 3x − 4y+ 8 = 0.

#### Solution

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

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 10 | Page 66