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 January 18th, 2014, 08:58 PM #1 Newbie   Joined: Dec 2013 Posts: 19 Thanks: 0 Geometrical locus I have changed the values so it's not solved for me but I have find the equation of the locus of point p which is equidistant from the y axis (point b)and point A (-3,1) So pa=pb Pa^2=pb^2 Pb^2= (x--3)^2 +(y-1)^2 Pb^2=x^2-6x+10+y^2-2y Then it equals the root of that? I am really banging my head against a wall on this one I have hunted back through my study notes and cannot even find a reference to solving a problem like this. January 19th, 2014, 04:29 AM #2 Newbie   Joined: Dec 2013 Posts: 19 Thanks: 0 Re: Geometrical locus i understand that whats its asking me for is an equation for a line thats equal distance from the 2 points and that ltiple points along that line.... aslong as it is equal distance from both A & B at any given point January 19th, 2014, 05:48 AM #3 Newbie   Joined: Dec 2013 Posts: 19 Thanks: 0 Re: Geometrical locus i think i nutted it out with my actual values given point A (3,-1) the point on the Y axis must be at (0,-1) PA=PB PA^2=PB^2 (x-0)^2+(y--1)^2=(x-3)^2+(y--1)^2 x^2+Y^2+2y+1=x^2-6x+9+y^2+2y+1 -6x+9=0 -6x=-9 x=9/6 x=3/2 therefore the line x=3/2 is the required equation for the loci between the 2 points as all points along the line will be equidistant between the 2 points. am i right? i still don't fully understand this topic given what i have been given to learn off Tags geometrical, locus Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mishin05 Calculus 0 December 18th, 2012 12:33 PM mrhesham Algebra 3 December 5th, 2010 11:51 AM kahalla Advanced Statistics 0 March 9th, 2010 01:24 AM grappler Algebra 1 June 11th, 2009 08:22 AM maly128 Algebra 3 March 13th, 2009 06:54 AM

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