# Find the Distance of the Point (3, 5) from the Line 2x + 3y = 14 Measured Parallel to the Line X − 2y = 1. - Mathematics

Find the distance of the point (3, 5) from the line 2x + 3y = 14 measured parallel to the line x − 2y = 1.

#### Solution

Here,

$\left( x_1 , y_1 \right) = A \left( 3, 5 \right)$

It is given that the required line is parallel to x − 2y = 1

$\Rightarrow 2y = x - 1$

$\Rightarrow y = \frac{1}{2}x - \frac{1}{2}$

$\therefore tan\theta = \frac{1}{2} \Rightarrow sin\theta = \frac{1}{\sqrt{5}}, cos\theta = \frac{2}{\sqrt{5}}$

So, the equation of the line is

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

$\Rightarrow \frac{x - 3}{\frac{2}{\sqrt{5}}} = \frac{y - 5}{\frac{1}{\sqrt{5}}}$

$\Rightarrow x - 3 = 2y - 10$

$\Rightarrow x - 2y + 7 = 0$

Let line $x - 2y + 7 = 0$ cut line 2x + 3y = 14 at P.

Let AP = r
Then, the coordinates of P are given by $\frac{x - 3}{\frac{2}{\sqrt{5}}} = \frac{y - 5}{\frac{1}{\sqrt{5}}} = r$ $\Rightarrow x = 3 + \frac{2r}{\sqrt{5}}, y = 5 + \frac{r}{\sqrt{5}}$

Thus, the coordinates of P are $\left( 3 + \frac{2r}{\sqrt{5}}, 5 + \frac{r}{\sqrt{5}} \right)$.

Clearly, P lies on the line 2x + 3y = 14.

$\therefore 2\left( 3 + \frac{2r}{\sqrt{5}} \right) + 3\left( 5 + \frac{r}{\sqrt{5}} \right) = 14$

$\Rightarrow 7 + \frac{7r}{\sqrt{5}} = 0$

$\Rightarrow r = - \sqrt{5}$

∴ AP =  $\left| r \right|$ = $\sqrt{5}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.8 | Q 9 | Page 66