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# Coordinate Geometry - Tough one

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Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12
Coordinate Geometry - Tough one [#permalink]

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20 Jun 2008, 15:04
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Q-coordinate.doc [23.5 KiB]

Director
Joined: 01 Jan 2008
Posts: 622
Re: Coordinate Geometry - Tough one [#permalink]

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20 Jun 2008, 15:15
1
KUDOS
product of slopes is equal to -1

(5/12)*(-5/b)=-1 -> b = 25/12
CEO
Joined: 17 May 2007
Posts: 2950
Re: Coordinate Geometry - Tough one [#permalink]

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20 Jun 2008, 15:53
1
KUDOS
Since its a right angle, both lines are perpendicular to each other. Which means that slope1 = -1/slope2

slope 1 = $$\frac{0-5}{-12-0}$$ = $$\frac{5}{12}$$

slope 2 = $$\frac{0-5}{b-0}$$ = $$\frac{-5}{b}$$

from the earlier formula $$\frac{5}{12}$$ = $$\frac{b}{5}$$
and b = $$\frac{25}{12}$$
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12
Re: Coordinate Geometry - Tough one [#permalink]

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20 Jun 2008, 15:33
maratikus wrote:
product of slopes is equal to -1

(5/12)*(-5/b)=-1 -> b = 25/12

what do you mean product of slopes=-1..can you elaborate?
Director
Joined: 27 May 2008
Posts: 543
Re: Coordinate Geometry - Tough one [#permalink]

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20 Jun 2008, 18:03
Alternate method

Its a right aggle triange....
(12+b)^2 = (12^2 + 5^2) + (b^2 + 5^2)
144 + b^2 + 24b = 144 + 25 + b^2 +25
24b=50
b=25/12
Manager
Joined: 08 Jun 2008
Posts: 71
Re: Coordinate Geometry - Tough one [#permalink]

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20 Jun 2008, 18:35
I took durgesh like approach too. I really liked maratikus' and bsd's approach. Its so easy and simple. I think this is the kind of thinking we need on the GMAT. I know about the product of the slopes but it never occured to me that I could solve that way. +1 to both of you maratikus & bsd.
Director
Joined: 01 May 2007
Posts: 793
Re: Coordinate Geometry - Tough one [#permalink]

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20 Jun 2008, 19:23
If you drop a line from a right angle to the opposite side so that it is perpendicular, it creates 3 similar triangles...so we can setup a proportion...

1) Notice we can solve the right triange...it is a 5,12,13 triangle...
2) Use the proportion to solve for b...

(12/5) = (5/b)...b = 25/12
Director
Joined: 14 Aug 2007
Posts: 728
Re: Coordinate Geometry - Tough one [#permalink]

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20 Jun 2008, 19:50
the product of the slopes of lines intersecting each other at 90 degrees angle = -1

the slope of one of the line is (5-0)/(0-(-12) = 5/12
so the slope of other line, intersecting this line at 90 degrees is -12/5 (for the product of slopes to be -1)

equation of this line then will be y= -12/5*x + C , y intercept C=5

b,0 lies on this line thus 0= -12/5*b + 5

b= 25/12
Re: Coordinate Geometry - Tough one   [#permalink] 20 Jun 2008, 19:50
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