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March 10th, 2013, 12:24 AM  #1 
Member Joined: Aug 2012 Posts: 71 Thanks: 1  coordinate geometry problem solving
I have absolutely no idea how to solve this. Your help would be very appreciated. A(5,3) and B(1,k) are two points. The perpendicular bisector of AB cuts the xaxis at 2. Find the two possible values for k. Many thanks. 
March 10th, 2013, 06:04 AM  #2 
Newbie Joined: Feb 2013 From: Australia Posts: 2 Thanks: 0  Re: coordinate geometry problem solving
Gradient of AB = (k  3)/(1  5) = (k + 3)/4 Gradient of any perpendicular to AB = negative reciprocal of this = 4/ (k + 3). Call this "m" Midpoint of AB = ((5 + 1)/2 , (3 + k)/2) = (3, (k3)/2) Call this (x1, y1) Equation of this line using y  y1 = m * (x  x1) is y  (k  3)/2 = 4/(k  3) * (x  3) And we're told the point this crosses the xaxis is (2, 0) so we substitute x = 2 and y = 0 0  (k  3)/2 = 4/(k  3) * (5) i.e. (3  k)/2 = 20/(k  3) i.e. (3  k)/2 = 20/(3  k) Crossmultiplying (3  k)^2 = 40 or same as (k  2)^2 = 40 taking square root k  2 = Sqrt40 or  sqrt 40 k  2 = 2 sqrt 10 or  2 sqrt 10 adding 2, k = 2 + 2sqrt 10 or 2  2sqrt 10 k = 8.32 or  4.32 
March 10th, 2013, 08:36 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1492 
Using a longwinded method makes careless errors easy to miss. For example, (3  k)² was changed to (k  2)² instead of (k  3)². However, there was an earlier mistake. The equation (3  k)/2 = 20/(k  3) should have been (3  k)/2 = 20/(k + 3); crossmultiplying then gives 3²  k² = 40, so k² = 49, i.e. k = ±7. Instead: as B and A are equidistant from (2, 0), 3² + k² = 7² + 3², and so k = ±7. 
March 10th, 2013, 10:33 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,672 Thanks: 741  Re: coordinate geometry problem solving Code: B(1,k) C(2,0) A(5,3) BC = SQRT[k^2 + (2  1)^2] = SQRT(k^2 + 9) AC = SQRT[(5  (2))^2 + (3  0)^2] = SQRT(54)**** Since BC = AC: k^2 + 9 = 54**** k^2 = 49 k = + 7 EDIT: **** should be 58 
March 10th, 2013, 12:31 PM  #5 
Member Joined: Aug 2012 Posts: 71 Thanks: 1  Re: coordinate geometry problem solving
Thanks everyone for the prompt replies.

March 10th, 2013, 02:47 PM  #6 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: coordinate geometry problem solving [color=#000000]Let C be the point (2,0), then . and .[/color] 
March 10th, 2013, 03:22 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1492 
In lengthening my explanation, Denis introduced a careless error in the arithmetic. I rest my face.

March 10th, 2013, 05:41 PM  #8  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,672 Thanks: 741  Re: Quote:
Edited; then recited a full 50beads rosary as penance...  

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