Coordinate Geometry

Question

The equations of two parallel lines are 2x+y-1=0 and 2x+y+11=0.Calculate the perpendicular distance between the lines

Answer 1

We have two parallel lines  2x+y-1=0   and   2x+y+11=0

Let's put those into slope intercept format y = mx +b

 

2x+y-1 = 0        2x+y+11=0

y = -2x+1          y = -2x-11

 

Both of these lines have slope m=-2

A line perpendicular to these would have a slope of 1/2 which is the negative reciprocal of -2

 

Pick a line with slope of 1/2  ....  Let's use y = 1/2(x)  or  2y=x

 

We need to find where this line intersects with the two other lines

Then find the distance between those two intersection points

 

Let's solve for intersection of 2y=x and y = -2x+1

 

2y = x           eqn 1

y = -2x +1    eqn 2

 

4y = 2x       2(eqn 1)

 y = -2x +1

-------------

5y = 1

 y = 1/5

 

Plug this back into one of the original equations..

2(1/5) = x

x = 2/5

These lines intersect at (2/5, 1/5)

 

Now solve for intersection of 2y=x and y = -2x -11

 

2y = x

 y = -2x -11

 

4y = 2x

  y = -2x -11

---------------

5y = -11

y = -11/5

 

2(-11/5) = x

x= -22/5

These two lines intersect at (-22/5, -11/5)

 

Since this line is perpendicular to our original 2 parallel lines the perpendicular distance

between the two parallel lines will be the distance from (2/5, 1/5) to (-22/5, -11/5)

The distance between two points is given by ...

d = √((x₂-x₁)²+(y₂-y₁)²)

d = √[((2/5 -(-22/5))² + (1/5-(-11/5))²]

d = √((24/5)²+ (12/5)²)

d= √(576/25 + 144/25)

d = √(720/25)

d = (12√5)/5

 

 

d ≅ 0.894 if you want a decimal answer

 

CORRECTION:  d ≅ 5.3666 in decimal form rounded to 4 places.  Thank you David W.