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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

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