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March 9th, 2013, 11:24 PM   #1
Doc
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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.
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March 10th, 2013, 05:04 AM   #2
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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
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March 10th, 2013, 07:36 AM   #3
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Using a long-winded method makes careless errors easy to miss.
For example, (3 - k) was changed to (k - 2) instead of (k - 3). 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 3 - k = -40, so k = 49, i.e. k = 7.

Instead: as B and A are equidistant from (-2, 0), 3 + k = 7 + 3, and so k = 7.
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March 10th, 2013, 09:33 AM   #4
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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
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March 10th, 2013, 11:31 AM   #5
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Re: coordinate geometry problem solving

Thanks everyone for the prompt replies.
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March 10th, 2013, 01:47 PM   #6
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Re: coordinate geometry problem solving

[color=#000000]Let C be the point (-2,0), then .





and .[/color]
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March 10th, 2013, 02:22 PM   #7
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In lengthening my explanation, Denis introduced a careless error in the arithmetic. I rest my face.
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March 10th, 2013, 04:41 PM   #8
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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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