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In the xy-coordinate system, if (m,n) and (m+2, n+k) are two

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In the xy-coordinate system, if (m,n) and (m+2, n+k) are two [#permalink]

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New post 03 Dec 2005, 07:07
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#1. In the xy-coordinate system, if (m,n) and (m+2, n+k) are two points on the line with the equation x=2y+5, then k=?

#2. In the xy-coordinate system, if (m,n) and (m+2, n+k) are two points on the line perpendicular to x=2y+5, then k=?

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Re: PS XY coordinate system [#permalink]

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New post 03 Dec 2005, 07:32
GMATT73 wrote:
#1. In the xy-coordinate system, if (m,n) and (m+2, n+k) are two points on the line with the equation x=2y+5, then k=?

#2. In the xy-coordinate system, if (m,n) and (m+2, n+k) are two points on the line perpendicular to x=2y+5, then k=?


the first one : we have the slope=(n+k-n)/(m+2-m) = k/2
x=2y+5 ---> y= 1/2x - 5/2 ---> k/2=1/2--->k=1

the secone one: the slope= k/2
the line perpendicular to x=2y+5 or y=1/2x-5/2 has
a slope= -1/ (1/2) =-2 --->k/2=-2 --> k=-4

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New post 06 Dec 2005, 22:32
If a point lies on the line, it has to satisfy the equation of the same line.

m=2n+5 and m+2=2(n+k)+5 :arrow: m+2=2n+2k+5 :arrow: 2=2k :arrow: k=1
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Re: PS XY coordinate system [#permalink]

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New post 07 Dec 2005, 05:46
[quote="laxieqvthe first one : we have the slope=(n+k-n)/(m+2-m) = k/2
x=2y+5 ---> y= 1/2x - 5/2 ---> k/2=1/2--->k=1

the secone one: the slope= k/2
the line perpendicular to x=2y+5 or y=1/2x-5/2 has
a slope= -1/ (1/2) =-2 --->k/2=-2 --> k=-4[/quote]

agreed.
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New post 07 Dec 2005, 21:59
OA is (1, -4). Piece of cake for you geniuses 8-)

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  [#permalink] 07 Dec 2005, 21:59
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In the xy-coordinate system, if (m,n) and (m+2, n+k) are two

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