# The distance of the point (3, 5) from the line \[2x + 3y - 14 = 0\] measured parallel to $x - 2y = 1$.

$

\left( a \right)\dfrac{7}{{\sqrt 5 }} \\

\left( b \right)\dfrac{7}{{\sqrt {13} }} \\

\left( c \right)\sqrt 5 \\

\left( d \right)\sqrt {13} \\

$

Last updated date: 27th Mar 2023

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Answer

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Hint: In this question, we use the parametric equation of straight line. Parametric form of the equation of a straight line: $\dfrac{{x - {x_1}}}{{\cos \theta }} = \dfrac{{y - {y_1}}}{{\sin \theta }} = r$ , where r is the distance between two points and $\theta $ is angle made by straight line with positive x-axis.

Complete step-by-step answer:

First we find slope of line passing through the point (3, 5) and parallel to $x - 2y = 1$

We know all parallel lines have the same slope so the slope of the line is $\dfrac{1}{2}$ .

Now, we can write as $\tan \theta = \dfrac{1}{2}$ .

We can find value of $\cos \theta $ and $\sin \theta $

$ \Rightarrow \cos \theta = \dfrac{2}{{\sqrt 5 }},\sin \theta = \dfrac{1}{{\sqrt 5 }}$

Let r be the required distance. Then the equation of line passing through (3,5) and parallel to $x - 2y = 1$.

Using a parametric equation of straight line .

$

\dfrac{{x - {x_1}}}{{\cos \theta }} = \dfrac{{y - {y_1}}}{{\sin \theta }} = r \\

\Rightarrow \dfrac{{x - 3}}{{\cos \theta }} = \dfrac{{y - 5}}{{\sin \theta }} = r \\

\Rightarrow x = 3 + r\cos \theta ,y = 5 + r\sin \theta \\

$

Since this point lies on the line 2x+3y−14=0 and also satisfies this line.

$ \Rightarrow 2\left( {3 + r\cos \theta } \right) + 3\left( {5 + r\sin \theta } \right) - 14 = 0$

Put the value of $\cos \theta $ and $\sin \theta $ .

$

\Rightarrow 2\left( {3 + \dfrac{{2r}}{{\sqrt 5 }}} \right) + 3\left( {5 + \dfrac{r}{{\sqrt 5 }}} \right) - 14 = 0 \\

\Rightarrow 6 + \dfrac{{4r}}{{\sqrt 5 }} + 15 + \dfrac{{3r}}{{\sqrt 5 }} - 14 = 0 \\

\Rightarrow 7 + \dfrac{{7r}}{{\sqrt 5 }} = 0 \\

\Rightarrow r = - \sqrt 5 \\

$

Required distance is $\sqrt 5 $ and negative signs represent only direction.

So, the correct option is (c).

Note: Whenever we face such types of problems we use some important points. Like we find an equation of line passing through a point and parallel to other lines by using a parametric form of straight line then after use given line we can get the required answer.

Complete step-by-step answer:

First we find slope of line passing through the point (3, 5) and parallel to $x - 2y = 1$

We know all parallel lines have the same slope so the slope of the line is $\dfrac{1}{2}$ .

Now, we can write as $\tan \theta = \dfrac{1}{2}$ .

We can find value of $\cos \theta $ and $\sin \theta $

$ \Rightarrow \cos \theta = \dfrac{2}{{\sqrt 5 }},\sin \theta = \dfrac{1}{{\sqrt 5 }}$

Let r be the required distance. Then the equation of line passing through (3,5) and parallel to $x - 2y = 1$.

Using a parametric equation of straight line .

$

\dfrac{{x - {x_1}}}{{\cos \theta }} = \dfrac{{y - {y_1}}}{{\sin \theta }} = r \\

\Rightarrow \dfrac{{x - 3}}{{\cos \theta }} = \dfrac{{y - 5}}{{\sin \theta }} = r \\

\Rightarrow x = 3 + r\cos \theta ,y = 5 + r\sin \theta \\

$

Since this point lies on the line 2x+3y−14=0 and also satisfies this line.

$ \Rightarrow 2\left( {3 + r\cos \theta } \right) + 3\left( {5 + r\sin \theta } \right) - 14 = 0$

Put the value of $\cos \theta $ and $\sin \theta $ .

$

\Rightarrow 2\left( {3 + \dfrac{{2r}}{{\sqrt 5 }}} \right) + 3\left( {5 + \dfrac{r}{{\sqrt 5 }}} \right) - 14 = 0 \\

\Rightarrow 6 + \dfrac{{4r}}{{\sqrt 5 }} + 15 + \dfrac{{3r}}{{\sqrt 5 }} - 14 = 0 \\

\Rightarrow 7 + \dfrac{{7r}}{{\sqrt 5 }} = 0 \\

\Rightarrow r = - \sqrt 5 \\

$

Required distance is $\sqrt 5 $ and negative signs represent only direction.

So, the correct option is (c).

Note: Whenever we face such types of problems we use some important points. Like we find an equation of line passing through a point and parallel to other lines by using a parametric form of straight line then after use given line we can get the required answer.

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