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PS-coordinates

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lan583
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PS-coordinates [#permalink] New post   12 Sep 2006, 05:14
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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positive soul
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Re: PS-coordinates [#permalink] New post   12 Sep 2006, 08:34
I don't know but it looks E to me. What's the logic to get to b?
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Re: PS-coordinates [#permalink] New post   12 Sep 2006, 14:03
to find the x coordinate we know that D is the midpoint between -8 and 20. so the distance between the 2 is 28. therefore the coordinate corresponding to this midpoint is either 20 -14 or -8+ 14 which both equals 6.
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Fig
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Re: PS-coordinates [#permalink] New post   Updated on: 13 Sep 2006, 06:13
My answer is (B).

As D is at ((20-8)/2,0), thus the answer is between (A) and (B).

I go on (B) because BD,on the graph, seems to be equal to AC

Originally posted by Fig on 12 Sep 2006, 14:30.
Last edited by Fig on 13 Sep 2006, 06:13, edited 1 time in total.
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haas_mba07
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Re: PS-coordinates [#permalink] New post   12 Sep 2006, 15:28
B. (6, 28)

Let B(x,y)

Given AB = BC

(x+8)^2 +y^2 = (x-20)^2 + y^2

Solving for x, x =6 (x^2 and y^2 cancel out)

Given AC = BC

or 28^2 = (20-6)^2 + y^2
or28^2 - 14^2 = y^2

or y^2 = (28+14)(28-14)

y^2 = 42x14 = 7x6x7x2 = 7^2 x 2^2 x 3

or y = 7x2x1.732

We know y is atleast 14. Eliminate A.

Therefore B(6,28).
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Re: PS-coordinates [#permalink] New post   12 Sep 2006, 19:43
ps_dahiya wrote:
positive soul wrote:
I don't know but it looks E to me. What's the logic to get to b?


Yes this must be E.
AC = BD = 28
AD = AC/2 = 14

what a silly mistake :wall



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Re: PS-coordinates [#permalink]
12 Sep 2006, 19:43
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