#
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/√13

b) √13 d) √5

1
Log in to add a comment

a) 7/√5 c)7/√13

b) √13 d) √5

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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

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