# The distance of the point (3,5) from the line 2x+3y-14=0 measured parallel to the line x-2y=1 is:a) 7/√5 c)7/√13b) √13 d) √5

1
by Deleted account

2014-10-03T09:50:02+05:30

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First find the line parallel to x-2y = 1 passing through (3,5)
It will be x-2y = c
put(3,5) in the line equation as it is a point on the line.
3- 2×5 = c
⇒c = -7
Equation of line : x - 2y = -7

Now find point of intersection of (x-2y=-7) and (2x+3y-14=0). You can find it by solving the 2 equations of line.

x - 2y + 7 = 0      <<<<<(1)
2x + 3y -14 = 0      <<<<<(2)
2x - 4y +14 = 0    <<<<<(3)(by multiplying 2 with eqn (1))
By Subtracting (3) from (2)
7y -28 =0
⇒ y = 4
Put y=4 in eqn(1)
x = 1

So the point of intersection is (1.4). Find the distance between (3,5) and (1.4).That is the required answer.
Distance = √(3-1)²+(5-4)² = √2²+1² = √5

And I am using my laptop.
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