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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= (x3)^2 +(y1)^2 Pb^2=x^26x+10+y^22y 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 (x0)^2+(y1)^2=(x3)^2+(y1)^2 x^2+Y^2+2y+1=x^26x+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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