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

The distance of the point (3, 5) from the line 2x+3y14=02x + 3y - 14 = 0 measured parallel to x2y=1x - 2y = 1.
(a)75 (b)713 (c)5 (d)13  \left( a \right)\dfrac{7}{{\sqrt 5 }} \\\ \left( b \right)\dfrac{7}{{\sqrt {13} }} \\\ \left( c \right)\sqrt 5 \\\ \left( d \right)\sqrt {13} \\\

Explanation

Solution

Hint: In this question, we use the parametric equation of straight line. Parametric form of the equation of a straight line: xx1cosθ=yy1sinθ=r\dfrac{{x - {x_1}}}{{\cos \theta }} = \dfrac{{y - {y_1}}}{{\sin \theta }} = r , where r is the distance between two points and θ\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 x2y=1x - 2y = 1
We know all parallel lines have the same slope so the slope of the line is 12\dfrac{1}{2} .
Now, we can write as tanθ=12\tan \theta = \dfrac{1}{2} .
We can find value of cosθ\cos \theta and sinθ\sin \theta
cosθ=25,sinθ=15\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 x2y=1x - 2y = 1.
Using a parametric equation of straight line .
xx1cosθ=yy1sinθ=r x3cosθ=y5sinθ=r x=3+rcosθ,y=5+rsinθ  \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.
2(3+rcosθ)+3(5+rsinθ)14=0\Rightarrow 2\left( {3 + r\cos \theta } \right) + 3\left( {5 + r\sin \theta } \right) - 14 = 0
Put the value of cosθ\cos \theta and sinθ\sin \theta .
2(3+2r5)+3(5+r5)14=0 6+4r5+15+3r514=0 7+7r5=0 r=5  \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 5\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.