Question
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+3y−14=0 measured parallel to x−2y=1.
(a)57 (b)137 (c)5 (d)13
Solution
Hint: In this question, we use the parametric equation of straight line. Parametric form of the equation of a straight line: cosθx−x1=sinθy−y1=r , where r is the distance between two points and θ 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 21 .
Now, we can write as tanθ=21 .
We can find value of cosθ and sinθ
⇒cosθ=52,sinθ=51
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 .
cosθx−x1=sinθy−y1=r ⇒cosθx−3=sinθy−5=r ⇒x=3+rcosθ,y=5+rsinθ
Since this point lies on the line 2x+3y−14=0 and also satisfies this line.
⇒2(3+rcosθ)+3(5+rsinθ)−14=0
Put the value of cosθ and sinθ .
⇒2(3+52r)+3(5+5r)−14=0 ⇒6+54r+15+53r−14=0 ⇒7+57r=0 ⇒r=−5
Required distance is 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.