Find the distance in units between the lines with equations 2x-5y=8 and y=2/5x-3

Question

A. (-23√29)/29

 

B. (7√29)/29

 

C. (-7√29)/29

 

D.  (23√29)/29

Answer 1

2x - 5y = 8  has slope 2/5

y = (2/5)x - 3 has slope 2/5

 

The lines are parallel.  We need to find a perpendicular line and look at the

intersection points of that perpendicular line with the two lines originally given.

Perpendicular lines have slopes that are negative reciprocals so we need a line

with slope m -1/(2/5) = -5/2

Let's just use:  y =(-5/2)x

 

y = (-5/2)x

2y = -5x

5x + 2y = 0

This line is perpendicular to the two lines given.  We need to solve two sets of simultaneous

equations to find where this line crosses the original lines, and look for the distance between

those two points

 

A)  Solve system for

     2x - 5y = 8

     5x + 2y = 0

    

    multiply 1st equation by 2 and 2nd by 5 and add

 

    4x - 10y = 16

   25x+10y = 0

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

   29x = 16

    x = 16/29

 

5x + 2y = 0

5(16/29) + 2y = 0

80/29 + 2y = 0

2y = -80/29

y = -40/29

 

Crossing point:  (16/29, -40/29)

 

B) Solve system for

    y = (2/5)x - 3

   5x + 2y = 0

 

   5x + 2[(2/5)x-3] = 0

   5x + (4/5)x - 6 = 0

   (29/5)x = 6

   x = 6(5/29)

   x = 30/29

 

  y = (2/5)x-3

  y = (2/5)(30/29) - 3

  y = 12/29 - 3

  y = 12/29 - 87/29

  y = -75/29

 

Crossing point:  (30/29, -75/29)

 

Find the distance between the two crossing points  (16/29, -40/29), (30/29, -75/29)

 

d = √[(x₂-x₁)²+(y₂-y₁)²]

 

d =  √[(30/29 - 16/29)² + (-75/29 - (-40/29))²]

 

d =  √[(14/29)² + (-35/29)²]

 

d =  √[196/29² + 1225/29²]

 

d = √1421/29

 

d =√[(49)(29)]/29

 

d = (7√29)/29