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Having Trouble Finding the Distance

  • Thread starter Lucretius
  • Start date
152
0
Summer has rolled around, but my studying continues. I want to be prepared for Calculus next year, so I bought some textbook at a thrift store. Already I'm running into problems! The book tells me to find the distance between P and L. P is a coordinate pair and L is a line equation. I have done three of these problems and all three have ended up being wrong, though I don't know where exactly I have made a mistake in my math.

[tex]P(-2,2)[/tex] [tex]L:2x+y=4[/tex]

I get the points for L by making x=0 and y=0, and I end up with:

[tex]P(-2,2)[/tex] [tex]L(2,4)[/tex]

Then I use the distance formula:

[tex]\sqrt{(2+2)^2+(4-2)^2}[/tex] which ultimately becomes [tex]\sqrt{20}[/tex], and finally [tex]2\sqrt{5}[/tex]

Unfortunately for me, my book reads: [tex]6\sqrt{5}[/tex]
 

jamesrc

Science Advisor
Gold Member
476
1
The first thing you need to realize is that the shortest distance between a point and a line is along a line segment perpendicular to the original line. This should allow you to write an equation for the line segment. See if that gets you going in the right direction; write back if you get stuck.
 
152
0
So, should the line equation be changed to [tex]y=2x-4[/tex]?

I tried making Line L into Line L' and then doing the distance formula but I get [tex]\sqrt{52}[/tex].
 

jamesrc

Science Advisor
Gold Member
476
1
No,

The equation of the original line is y = 4-2x
What is the slope of a line that is perpendicular to this line? The slope of a perpendicular line will be the negative reciprocal slope of the original line.
Then you should set up an expression for the slope of the new line (which you now know). Something like m = (y1-y2)/(x1-x2) The subscript 2 could be your ordered pair. The subscript 1 would then be the point on the original line that you want to find.
I hope that helps.
 
152
0
I'll try to do it tomorrow. I must be missing something right in front of my eyes. After 55 minutes of staring at the page, I just draw blanks.
 
152
0
Okay, I am about to go insane, no matter what I try, I cannot get this problem. I spent an hour looking at it last night, and an hour today, and I've tried everything — even what the book is telling me is not helping, and I cannot get the answer they tell me.

I found the line perpendicular to line L running through point P, which is (-2,2)
The equation of the line is x-2y=-6, or y=3+1/2x. The line it is perpendicular to is y=-2x+4.

The book I use tells me to set these equations equal to eachother, and so I have: 3+1/2x=-2x+4, which leads to 2 1/2x = 1, and I get 4/10 for x. I then get y by plugging 4/10 in the first line equation. -2(4/10)+4 = 3 2/10.

Using the distance formula I get: [tex]\sqrt{(-2-\frac{4}{10})+(2-3\frac{2}{10})[/tex], which becomes [tex]\sqrt{\frac{832}{100}[/tex]. This does not become what the book says the answer is: [tex] \frac{6}{\sqrt{5}}[/tex]
 

FredGarvin

Science Advisor
5,050
6
Lucretius said:
Summer has rolled around, but my studying continues. I want to be prepared for Calculus next year, so I bought some textbook at a thrift store. Already I'm running into problems! The book tells me to find the distance between P and L. P is a coordinate pair and L is a line equation. I have done three of these problems and all three have ended up being wrong, though I don't know where exactly I have made a mistake in my math.

[tex]P(-2,2)[/tex] [tex]L:2x+y=4[/tex]

I get the points for L by making x=0 and y=0, and I end up with:

[tex]P(-2,2)[/tex] [tex]L(2,4)[/tex]

Then I use the distance formula:

[tex]\sqrt{(2+2)^2+(4-2)^2}[/tex] which ultimately becomes [tex]\sqrt{20}[/tex], and finally [tex]2\sqrt{5}[/tex]

Unfortunately for me, my book reads: [tex]6\sqrt{5}[/tex]
Do you know the point-slope equation for a line? It looks like this:
[tex]y-y_1 = m(x-x_1)[/tex]

You need to develop the equation for the line that is perpendicular to the line you have been given, L, and through the point given, P.

The slope of a perpendicular line is the negative of the reciprocal, or [tex]M_{\bot} = -(\frac{1}{m})[/tex]

You can use the point and the perpendicular slope to come up with the equation of the perpendicular line:
[tex]y-y_1 = m(x-x_1)[/tex]
[tex]y-2 = \frac{1}{2}(x+2)[/tex]
[tex]y = \frac{1}{2} x + 3[/tex]

Now that you have the 2 line equations (y=-2x+4 and y=1/2x+3), set them equal to each other to calculate one of the variables. I'll set the Y variables equal to calculate x:
[tex]\frac{1}{2} x + 3 = -2x + 4[/tex]
[tex] x = \frac{2}{5}[/tex]

Now use the value of x calculated to calculate y by using eiter equation:
[tex]y = -2x + 4[/tex]
[tex]y = -2(\frac{2}{5}) + 4[/tex]
[tex]y = \frac{16}{5}[/tex]

Now you have two points to plug and chug on the distance formula:
[tex]d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]

[tex]d = \sqrt{(-2-\frac{2}{5})^2 + (2-\frac{16}{5})^2}[/tex]

Which ends up to equal [tex]\frac{6 \sqrt{5}}{5}[/tex]

I double checked the answer on my CAD station so I know it's right and your book answer has a type-o.
 
31
0
my advice is to graph it all, then see if it sparks any sudden inspiration
 

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