1. ## Line Distance?

3x-4y-10=0
6x-8y+5=0

are two parallel lines. How to calculate the distance between them.

The formula for calculating the distance between a line and a point is
[code]
point A(a,b)
line p = Ax+By+C=0

distance d = (|Aa+Bb+C|) / (squareroot(A^2+B^2)

Hoe do i get a point from a line. I know it was very easy but i forgot it.

Thanks!  Reply With Quote

2. your equations can also be represented as
line 1: y = 3x/4 -10/4
line 2: y = 6x/8 +5/8 = 3x/4 +5/8

The 3/4 is the slope of the lines, a line that crosses both lines with the shortest distance has a slope of -4/3

using the new line 3: y = -4x/3 we can now find the intercept points for lines 1 and 2 with line 3

3x/4 - 10/4 = -4x/3
9x-30 = -16x
x = 6/5
y = 3/4 * 6/5 - 10/4 = 18/20 - 50/20 = -32/20 = -8/5

3x/4 + 5/8 = -4x/3
18x+15 = -32x
x = -3/8
y = 3/4 * 3/8 + 5/8 = 9/32 + 20/32 = 29/32

Point 1 = (6/5, -8/5)
Point 2 = (-3/8, 29/32)

Then the ditance between these two points is

sqrt((6/5+3/8)^2+(-8/5-29/32)^2) = 2.96...

This is the shortest distance bewtween your two lines.  Reply With Quote

3. Well illuminator..

Your Idea of Introducing a Third perpendicular line was
good but not simple enough.

Proof :
You made a simple calculation mistake /* second part */
3x/4 + 5/8 = -4x/3
18x+15 = -32x
x = -3/8
y = 3/4 * 3/8 + 5/8 = 9/32 + 20/32 = 29/32
3x/4 + 5/8 = -4x/3

(6x+5)/8 = -4x /3

3(6x+5)=-32x

or x=-3/10 (not -3/8)

therfore y=2/5

and the two points are (6/5,-8/5) and (-3/10,2/5)

and distance between these two points is 5/2 = 2.5

----
Here is a much simpler way...

As we see from the equations

3x-4y-10=0
6x-8y+5=0

are not parallel to the x -axis or y-axis

Therefore they intersect both the axes at some points.

The first equation for example Intersects x-axis at
(10/3,0)
how ??
The y co-ordinate at the point the line cuts x-axis is zero.
put y=0 in the equation

3x-10=0 ==> x = 10/3 and y=0
therefore we found one point on the line 3x-4y-10=0

you got the the other line equation 6x-8y+5 =0

a,b -> 10/3,0

d = (|Aa+Bb+C|) / (squareroot(A^2+B^2)

d = |(6*10/3)+(-8*0)+5| / Sqr((6^2+(-8)^2))

d = 2.5

--------
You can also do it the otherway..
Similarly you can take a point on the second line
6x-8y+5=0 and find perpencicular distance to the other line
3x-4y-10=0

at the point of intersection line 6x-8y+5=0 with x-axis, y=0
6x-8(0)+5=0

==> x = -5/6 and y=0

therfore use you perpendicular distance formula..
(a,b) -> (-5/6,0)

d = |3*(-5/6)+(-4*0)-10| / Sqr(3^2+(-4)^2)
d = (25/2)/5
d = 2.5

---  Reply With Quote

4. Thanks for the replies guys! I got it!  Reply With Quote

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