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 March 9th, 2013, 11:24 PM #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 x-axis at -2. Find the two possible values for k. Many thanks.
 March 10th, 2013, 05:04 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, (k-3)/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 x-axis 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) Cross-multiplying (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, 07:36 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Using a long-winded 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); cross-multiplying 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, 09:33 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,612 Thanks: 845 Re: coordinate geometry problem solving Code: B(1,k) C(-2,0) A(5,-3) That solution had me confused too! Same idea as Skip's: 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, 11:31 AM #5 Member   Joined: Aug 2012 Posts: 71 Thanks: 1 Re: coordinate geometry problem solving Thanks everyone for the prompt replies.
 March 10th, 2013, 01:47 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 $M=\frac{A+B}{2}=\left(3,\frac{k-3}{2}\right)$. $\vec{BM}=\left<2,\frac{k-3}{2}-k\right=>=\left<2,-\frac{k+3}{2}\right=>=$ $\vec{CM}=\left<5,\frac{k-3}{2}\right=>=$ and $\vec{BM}\cdot \vec{CM}=0\Leftrightarrow 10-\frac{(k-3)(k+3)}{4}=0\Leftrightarrow k^2=49\Leftrightarrow k=\pm 7$.[/color]
 March 10th, 2013, 02:22 PM #7 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 In lengthening my explanation, Denis introduced a careless error in the arithmetic. I rest my face.
March 10th, 2013, 04:41 PM   #8
Math Team

Joined: Oct 2011

Posts: 12,612
Thanks: 845

Re:

Quote:
 Originally Posted by skipjack In lengthening my explanation, Denis introduced a careless error in the arithmetic. I rest my face.
Yikes...49 + ..1 finger,2 fingers... 9 fingers = 58.
Edited; then recited a full 50beads rosary as penance...

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