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November 5th, 2009, 04:01 PM   #1
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Analytic Geometry

The y-intercept of a line is 8 and the distance from the line to (1,4) is -4. Find its equation.

I have no idea how to do this!!!! HELP
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November 5th, 2009, 04:51 PM   #2
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Re: Analytic Geometry

2 things:
1. Do you mean the shortest distance?
2. What do you mean by a distance of "-4"?
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November 5th, 2009, 06:01 PM   #3
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Re: Analytic Geometry

NOTE: I'm just guessing this since I am presently doped up on rather strong pain pills [really], and just passing a little time before hitting the sack a bit earlier than usual. I hope it helps get you on the right track and not a wrong one.

1. Write the general equation for the pencil of lines with y-intercept of 8.
2. Write the squared distance formula from a general point (x,y) on the line to the point (1,4) and the distance 4 [squared]. That's a circle, a quadratic in x and y.
3. Equate the negative inverse of the general slope, "m" to the slope of the line through (x,y) and the point (1,4).
4. Use #3 and #1 to equate the "m"s and so have a second quadratic equation in x and y.
5. Solve the system involving the equations found in #2 and #4.

That should give second point [or two?] on the line. Using two points, you have the equation.

I won't mind at all if someone shoots holes through this or comes up with a much better approach to this problem.
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November 6th, 2009, 11:00 AM   #4
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Re: Analytic Geometry

Quote:
Originally Posted by amkroes
The y-intercept of a line is 8 and the distance from the line to (1,4) is -4. Find its equation.

I have no idea how to do this!!!! HELP
I suggest you draw a graph and plot the points (0, and (1,4). Now, draw a circle with (1,4) being the center, and your radius being 4. I'm not sure what you mean by distance being -4. Maybe that's the closest the line gets to the point (1,4)? And how do you know it's a negative distance? What point of view are you talking about? That circle you just drew around (1,4) has an infinite amount of points that have a distance of 4 around (1,4).
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November 6th, 2009, 11:30 AM   #5
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Re: Analytic Geometry

A line drawn from (1, 4), to a line through (0, 8) that is 4 units from (1, 4), must be perpendicular to the line through (0, 8). Using this you can construct a right angled triangle with sides 1, 4 and ?(17), and vertexes (0, 8) and (1, 4). Use trigonometry to find the coordinates of the third vertex and then use that to find the equation of the line.

Note: Two lines may be constructed passing through (0, 8) and 4 units (shortest distance) from (1, 4).
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November 6th, 2009, 11:56 AM   #6
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Re: Analytic Geometry

Stubbornly following my own procedure, easily solving the two quadratics, I got (1, as one other possible point on the line. That seems to fit nicely, and the line is then y = 8.
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November 6th, 2009, 12:06 PM   #7
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Re: Analytic Geometry

Quote:
Originally Posted by David
Stubbornly following my own procedure, easily solving the two quadratics, I got (1, as one other possible point on the line. That seems to fit nicely, and the line is then y = 8.
Yes, y = 8, and your method, that works. I got (-15/17, 128,17) for the other point, thus y = 8/15 x + 8 as an equation of the line, which, I believe, is also correct. Not purely an analytic geometry approach (mine) though, I suppose.
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November 6th, 2009, 04:54 PM   #8
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Re: Analytic Geometry

Quote:
Originally Posted by greg1313
Quote:
Originally Posted by David
Stubbornly following my own procedure, easily solving the two quadratics, I got (1, as one other possible point on the line. That seems to fit nicely, and the line is then y = 8.
Yes, y = 8, and your method, that works. I got (-15/17, 128,17) for the other point, thus y = 8/15 x + 8 as an equation of the line, which, I believe, is also correct. Not purely an analytic geometry approach (mine) though, I suppose.
There are two solutions to the quadratic. I got a bit lazy and just gave the one.
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November 6th, 2009, 05:46 PM   #9
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Re: Analytic Geometry

Quote:
Originally Posted by David
There are two solutions to the quadratic. I got a bit lazy and just gave the one.
Yes, I saw that when I tried your method. Neat!
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November 6th, 2009, 06:41 PM   #10
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Re: Analytic Geometry

Quote:
Originally Posted by greg1313
Quote:
Originally Posted by David
There are two solutions to the quadratic. I got a bit lazy and just gave the one.
Yes, I saw that when I tried your method. Neat!
Thanks, but what I think is neat is the people who think up these problems and the way in this one that the squared x and y disappear to leave a linear relation in x and y to sub back in. My idea of fun, I guess. As I read through the solutions to other problems given by others here such as yourself and skipjack in particular, I am amazed at the extent of their knowledge and willingness to help others. It really is a great forum, and I'd like to see it kept that way. I'll be leaving soon. No offense to anyone and it's been fun. I'm just getting bit too old and tired for the game, and there are other, better really, things to do in this life like play with grandchildren.

Ciao.
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