SOLUTION: Find the distance between two parallel lines having the following equations: y=3/8x+4 y=3/8x-5

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Question 1150216: Find the distance between two parallel lines having the following equations: y=3/8x+4 y=3/8x-5
Found 2 solutions by Edwin McCravy, Alan3354:
Answer by Edwin McCravy(20067) About Me  (Show Source):
You can put this solution on YOUR website!
Get one line in the form Ax+By+C=0

Find any point on the other line.

Use this formula:

The perpendicular distance from the point (x1,y1)
to the line Ax+By+C=0 is

d=abs%28Ax%5B1%5D%2BBy%5B1%5D%2BC%29%2Fsqrt%28A%5E2%2BB%5E2%29

y=3/8x+4
Multiply through by 8
8y=3x+32
0=3x-8y+32
3x-8y+32=0

A=3, B=-8, C=32

Pick any point on y=3/8x-5, say its y-intercept (0,-5)

d=abs%283%280%29-8%28-5%29%2B32%29%2Fsqrt%283%5E2%2B%28-8%29%5E2%29

d=abs%280%2B40%2B32%29%2Fsqrt%289%2B64%29

d=abs%2872%29%2Fsqrt%2873%29

d=72%2Fsqrt%2873%29

Rationalize the denominator:

d=72sqrt%2873%29%2F73

Edwin

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the distance between two parallel lines having the following equations: y=3/8x+4 y=3/8x-5
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Assuming you mean y = (3/8)x + 4 and y = (3/8)x - 5
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The distance between the y-intercepts is 9.
The angles of both lines wrt the x-axis is atan(3/8)
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The angle between the line perpendicular to the given lines and y-axis is also atan(3/8)
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distance between the given lines = 9*cos(atan(3/8)) = ~ 8.427 units