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In the attached figure: On the xy-coordinate plane, a

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In the attached figure: On the xy-coordinate plane, a [#permalink]

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New post 24 Aug 2011, 19:25
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In the attached figure:
On the xy-coordinate plane, a triangular region is bounded by the lines y = 3, x = -6, and y = cx + d. One vertex of this region is (-6,0). What is the perimeter of this region?

(1) d = 5

(2) c = 5/6

For the 1st option, I did get y=5/6x + 5. And the intersection this line and y =3 is point -12/5,3. But I did not get how the distance between (-6,3) and (-12/5, 3) is 3/5

Can anyone please explain?

Thanks
NAD
[Reveal] Spoiler: OA

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Re: Perimeter of region bound by Lines [#permalink]

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New post 25 Aug 2011, 00:44
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nades09 wrote:
In the attached figure:
On the xy-coordinate plane, a triangular region is bounded by the lines y = 3, x = -6, and y = cx + d. One vertex of this region is (-6,0). What is the perimeter of this region?

(1) d = 5

(2) c = 5/6

For the 1st option, I did get y=5/6x + 5. And the intersection this line and y =3 is point -12/5,3. But I did not get how the distance between (-6,3) and (-12/5, 3) is 3/5

Can anyone please explain?

Thanks
NAD


That is an error. The distance between (-6, 3) and (-12/5, 3) is not 3/5. It is 18/5 or 3(3/5).

I suggest that you should not get into actual solving. Just use logic to figure out whether data is sufficient or not.

To get the perimeter of the triangle, you need to know the length of the sides. To know the length of the sides, you need the co-ordinates of the vertices. If you know the two vertices (x1, y1) and (x2, y2), the length of the side is given by \(\sqrt{(x1-x2)^2 + (y1-y2)^2}\). To know the co-ordinates of the vertices, you need to know the equation of the third line, y = cx + d. When you know the equation of two lines, you can find their point of intersection by solving them simultaneously.

If the equation of the third line is represented by y = cx + d, c is the slope of the line and d is the y intercept. As seen in the figure, this line passes through (-6, 0). All that is missing to get the perimeter is the equation of this line.

Statement 1 tells us that the y intercept is 5. This means that this line passes through (0, 5). So now we know 2 points through which this line passes (-6, 0) and (0, 5). Hence we can find the equation of the line. Sufficient.

Statement 2 tells us that the slope of the line is 5/6. Now we know a point through which this line passes (-6, 0) and its slope. This is sufficient to find the equation of the line. Sufficient.

Answer D

For more on equations of lines, y intercept, points etc, check:
http://www.veritasprep.com/blog/2010/12 ... he-graphs/
http://www.veritasprep.com/blog/2010/12 ... s-part-ii/
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Re: Perimeter of region bound by Lines [#permalink]

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New post 25 Aug 2011, 18:09
Thanks Karishma for the shortcut solution. I was also getting 3(3/5). I did get the explanation that was provided as the solution, but was not sure about 3/5 and two people had said the same. The links explaining coordinate geometry are very useful. Thanks!

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On the xy-coordinate plane, a triangular region is bounded by the line [#permalink]

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New post 12 Aug 2015, 02:27
On the xy-coordinate plane, a triangular region is bounded by the lines y = 3, x = -6, and y = cx + d. One vertex of this region is (-6,0). What is the perimeter of this region?

(1) d = 5

(2) c = 5/6

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Re: In the attached figure: On the xy-coordinate plane, a [#permalink]

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New post 12 Aug 2015, 06:06
PathFinder007 wrote:
On the xy-coordinate plane, a triangular region is bounded by the lines y = 3, x = -6, and y = cx + d. One vertex of this region is (-6,0). What is the perimeter of this region?

(1) d = 5

(2) c = 5/6


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Kudos [?]: 135629 [0], given: 12705

Re: In the attached figure: On the xy-coordinate plane, a   [#permalink] 12 Aug 2015, 06:06
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