finding distances between pairs of parallel lines

cole92

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Mar 30, 2006
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Ok well, I don't have a very good math teacher at all, so I find myself basically teaching myself math by reading in the book (which we aren't allowed to bring home because we don't have enough). So anyway, please help me on this and any help at all will be greatly appreciated:

1) Find the distance between each pair of parallel lines:

y = x + 9
y = x + 3

2) Find the distance between each pair of parallel lines:

y = -5x
y = -5x + 26

I have other problems, but once I understand how to do these two, I think I can figure out how to do them...Thanks in advance for any help.
 

jonboy

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Jun 8, 2006
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Hi Cole92!

A good way to do this is to graph this equations and improvise a way to solve it.

Here's the graph:

60934546gl8.jpg


So count how many units up the lower line is from the upper line (look at the y - axis). Or you could use the distance formula knowing your points since the slope and y - intercept is constant.

You can derive the points \(\displaystyle \{(0,3)\,,\,(0,9)\}\) because they are you y - intercepts.

So now use the distance formula:\(\displaystyle \L \;d\,=\,\sqrt{{(x_1\,-\,x_2)}^2\,+\,{(y_1\,-\,y_2)}^2}\)

....To get \(\displaystyle \L \;d\,=\,6\)

Think you can do the second one now?
 

tkhunny

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Except that distance normally would be defined at PERPENDICULAR distance.

You might be shooting for \(\displaystyle 3\sqrt{2}\) and \(\displaystyle \sqrt{26}\).

Read your definition of "Distance" and proceed according to that definition.
 

jonboy

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tkhunny said:
Except that distance normally would be defined at PERPENDICULAR distance.

You might be shooting for \(\displaystyle 3\sqrt{2}\) and \(\displaystyle \sqrt{26}\).

Read your definition of "Distance" and proceed according to that definition.

How did you get those answers?
 

cole92

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Mar 30, 2006
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Yes, I believe it is supposed to be perpendicular.

Ok, from what I am understanding, this is how I did number two...

y = -5x ----->slope is -5
y(2) = -5x + 26 ------> made slope 1/5

then...

I made the equation y(2) = 1/5x + 26 ...I get the points (0,26)

Then I plugged in:

-5x = 1/5x + 26
solve...
x = -5 [gives me points (-5,?)]

Finally I plugged in -5 (my answer for x) into the first equation and got 25 [making my second set of points (-5,25)]

Then I did the distance formula using points (0,26) and (-5,25)...and my answer was:

Square Root of 26?

(I'm sorry if this is wrong. I'm really confused by my teacher and am trying to put my notes with what you guys are showing me).
 

pka

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Jan 29, 2005
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Given \(\displaystyle T:(p,q)\quad \& \quad Ax + By + C = 0\), a point and a line, then the distance from the point to the line is given by: \(\displaystyle D = \frac{{\left| {Ap + Bq + C} \right|}}{{\sqrt {A^2 + B^2 } }}.\)

Now with two parallel lines, take any point on one of the lines then find its distance to the other line.
 

tkhunny

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y = x + 9
y = x + 3 ==> x - y + 3 = 0

Pick ANY point on the first, (0,9).
Calculate the distance to the other \(\displaystyle \frac{|1(0)-1(9)+3|}{\sqrt{(1)^{2}+(-1)^{2}}}\;=\;\frac{6}{\sqrt{2}}\;=\;3\sqrt{2}\)

y = -5x
y = -5x + 26 ==> 5x + y - 26 = 0

This one's even easier, since it contains the Origin. Pick (0,0)
Calculate the distance to the other \(\displaystyle \frac{|5(0)+1(0)-26|}{\sqrt{(5)^{2}+(1)^{2}}}\;=\;\frac{26}{\sqrt{26}}\;=\;\sqrt{26}\)

That's about it.
 

cole92

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Mar 30, 2006
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Great so then my answer to number two is correct. Ok guys thanks for all of your help...I think I get it now.
 

cole92

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Mar 30, 2006
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65
One last question, I think I totally screwed this one up:

y = -x - 7
7 = -x - 9
 

skeeter

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cole92 said:
One last question, I think I totally screwed this one up:

y = -x - 7
7 = -x - 9

did you mean for the second equation to be y = -x - 9 ?

if so, the perpendicular distance between the lines is \(\displaystyle \L \sqrt{2}\).
 

cole92

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Mar 30, 2006
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65
Yes, sorry about that. Ok, well that is what I got, but I also got a way different answer. By any chance could you show me what you did so I can compare it with mine...cause I don't think I got it the right way(not to mention it took me a really long time).
 

skeeter

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Dec 15, 2005
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I sketched the graphs and used geometry ...

the distance between the y-intercepts is 2, and that distance is the hypotenuse of a right isosceles triangle whose vertices are the two y-intercepts and the point of intersection of a perpendicular drawn from (0,-7) to the line y = -x - 9.

The perpendicular distance is the short leg of the aforementioned triangle.
 
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