November 5th, 2009, 04:01 PM  #1 
Newbie Joined: Nov 2009 Posts: 1 Thanks: 0  Analytic Geometry
The yintercept 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 yintercept 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:
 
November 6th, 2009, 11:30 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 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: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond  Re: Analytic Geometry Quote:
 
November 6th, 2009, 04:54 PM  #8  
Senior Member Joined: Mar 2007 Posts: 428 Thanks: 0  Re: Analytic Geometry Quote:
 
November 6th, 2009, 05:46 PM  #9  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond  Re: Analytic Geometry Quote:
 
November 6th, 2009, 06:41 PM  #10  
Senior Member Joined: Mar 2007 Posts: 428 Thanks: 0  Re: Analytic Geometry Quote:
Ciao.  

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