My Math Forum Analytic Geometry

 Geometry Geometry Math Forum

 November 5th, 2009, 04:01 PM #1 Newbie   Joined: Nov 2009 Posts: 1 Thanks: 0 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
 November 5th, 2009, 04:51 PM #2 Senior Member   Joined: May 2008 From: Sacramento, California Posts: 299 Thanks: 0 Re: Analytic Geometry 2 things: 1. Do you mean the shortest distance? 2. What do you mean by a distance of "-4"?
 November 5th, 2009, 06:01 PM #3 Senior Member   Joined: Mar 2007 Posts: 428 Thanks: 0 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.
November 6th, 2009, 11:00 AM   #4
Senior Member

Joined: Oct 2009

Posts: 105
Thanks: 0

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).

 November 6th, 2009, 11:30 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond 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).
 November 6th, 2009, 11:56 AM #6 Senior Member   Joined: Mar 2007 Posts: 428 Thanks: 0 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.
November 6th, 2009, 12:06 PM   #7
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,963
Thanks: 1148

Math Focus: Elementary mathematics and beyond
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.

November 6th, 2009, 04:54 PM   #8
Senior Member

Joined: Mar 2007

Posts: 428
Thanks: 0

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.

November 6th, 2009, 05:46 PM   #9
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,963
Thanks: 1148

Math Focus: Elementary mathematics and beyond
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!

November 6th, 2009, 06:41 PM   #10
Senior Member

Joined: Mar 2007

Posts: 428
Thanks: 0

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.

 Tags analytic, geometry

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post condemath Geometry 1 August 29th, 2011 03:50 AM Rayne Geometry 0 November 2nd, 2010 06:32 AM Rayne Geometry 0 November 2nd, 2010 05:57 AM Rayne Geometry 0 November 2nd, 2010 05:49 AM nortpron Geometry 6 April 5th, 2010 01:07 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top