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Given line L, and a parallel line that runs through point

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Given line L, and a parallel line that runs through point  [#permalink]

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New post Updated on: 13 Apr 2013, 04:27
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Given line L (illustrated in graph), and a parallel line that runs through point (-1,5), what is the perimeter of a rectangle whose sides run along the two lines and has a length of 9?

A. 28
B. 9*(15)^0.5
C. 18+2*(20)^0.5
D. 36
E. 45

Originally posted by 12bhang on 13 Apr 2013, 04:16.
Last edited by Bunuel on 13 Apr 2013, 04:27, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Given line L, and a parallel line that runs through point  [#permalink]

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New post 13 Apr 2013, 05:36
9
1
Line:
y = (3/4)x - 1/2

point:
(-1,5)

distance of a point (p,q) from a line ax+by+c = 0 is given by formula:

\(d= |(ap + bq + c)/sqrt(a^2 + b^2)|\)

Simplify the equation:

Multiply 4 in each side:
4y = 3x - 2
=> 3x -4y - 2 = 0

Use the formula:
\(d=|{ 3*(-1) - 4*5 - 2}/ sqrt(3^2 + (-4)^2)|
= |(-3 -20 - 2)/5| = |-25/5| = 5\)

Perimeter of the recatangle = 2(length + breadth)
length = 9
breadth = distance of the point from the line = 5

Perimeter = 2*( 9 +5) = 3*14 = 28
Option A.
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Re: Given line L, and a parallel line that runs through point  [#permalink]

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New post 13 Apr 2013, 04:57
2
The parallel line is found using the formula y-y0=m(x-x0) , y-5=3/4(x+1)
The question comes down to what is the distance between \(y=\frac{3}{4}x-\frac{1}{2}\) and \(y=\frac{3}{4}x+\frac{23}{4}\)?
I was not able to find a quick and easy solution to this question, so I took the perpendicular line \(y=-4/3x\) and calculated the intersections.
3/4x+23/4=-4/3x point\((-\frac{69}{25},\frac{92}{25})\)
3/4x-1/2=-4/3x pont\((\frac{6}{25},-\frac{8}{25})\)
Now using Pitagora we must find the hypotenuse of this triangle which has lengths \(\frac{92+8}{25}=4\) and \(\frac{69+6}{25}=3\) (refer to the picture and sorry for the bad quality...)
So hypotenuse = 5 and perimeter = 9+9+5+5=28
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Re: Given line L, and a parallel line that runs through point  [#permalink]

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New post 15 Apr 2013, 06:19
4
12bhang wrote:
Attachment:
CG.png
Given line L (illustrated in graph), and a parallel line that runs through point (-1,5), what is the perimeter of a rectangle whose sides run along the two lines and has a length of 9?

A. 28
B. 9*(15)^0.5
C. 18+2*(20)^0.5
D. 36
E. 45


The question asks nothing more than calculating the perpendicular distance between the point (-1,5) and the line 3x-4y-2=0. This distance = \(|3(-1)-4(5)-2|/\sqrt{3^2+4^2}\)= |-25|/5 = 5. Thus the perimeter is 2*(5+9) = 28.
A.
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Re: Given line L, and a parallel line that runs through point  [#permalink]

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New post 15 Apr 2013, 12:12
12bhang wrote:
Attachment:
CG.png
Given line L (illustrated in graph), and a parallel line that runs through point (-1,5), what is the perimeter of a rectangle whose sides run along the two lines and has a length of 9?

A. 28
B. 9*(15)^0.5
C. 18+2*(20)^0.5
D. 36
E. 45


Another way to work out....
The equation of line parallel to y=3/4x-1/2 passing through -1,5 will be
y=3/4x+23/4
Now distance between two parallel lines y=mx+c and y=mx+c1 is given by
= |c1-c| / sq rt(m^2+1)

here we have |23/4 + 1/2| /sq root (3/4^2 + 1) which comes out to be 5.
Perimeter is 9*2 + 5*2 = 28
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Re: Given line L, and a parallel line that runs through point  [#permalink]

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New post 27 Apr 2013, 22:01
12bhang wrote:
Attachment:
CG.png
Given line L (illustrated in graph), and a parallel line that runs through point (-1,5), what is the perimeter of a rectangle whose sides run along the two lines and has a length of 9?

A. 28
B. 9*(15)^0.5
C. 18+2*(20)^0.5
D. 36
E. 45




The slope will be same for parallel lines. Hence, we can write the equation of the line which passes through point (-1.5) as y = 3/4x+b.
After substituting the values of x&y, the equation will become,
5=3/4*-1 + b
5 = -3/4 + b
b = 23/4

The question already provides the length of one side and which is equal to 9. Now, we have to find the length of the other side.

In other words, it requires us to find the distance b/w two parallel lines, y = 3/4x - 1/2 & y = 3/4x + 23/4.

The formula to find the distance b/w two parallel lines is |b-c| / √(m²+1)

|23/4 + 1/2| / √(3/4)²+1) = (25/4) / √(25/16) = 5

Hence, the perimeter of the rectangle in the xy plane will become = 2 (9 + 5) = 2*14 = 28
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Re: Given line L, and a parallel line that runs through point  [#permalink]

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New post 14 Apr 2014, 05:17
Here's the way I solved this one.

We basically have a rectangle and we need to find the perimeter 2 (L+W). We know the length is 9, so we need to find the width which is equal to the distance between the two parallel lines

Thus we know that the other line running through (-1,5) is y=3/4x+b

Distance between two parallel lines is 25/4 / sqrt (9/16+1) = 5

Therefore perimeter is thus 28

Hope this helps
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Re: Given line L, and a parallel line that runs through point  [#permalink]

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New post 14 Apr 2014, 22:48
Slope of perpendicular line to 3/4x-1/2 is equal to -4/3 and this means that we have two legs 4 and 3 and hypotenuse 5 which is side of rectangle, so perimeter is 2*9+2*5=28. Answer is A

Problem is that slope -4/3 can result in hypotenuse equal to 5,10, 15, 20 etc. But only option that fits to answer choices is 5
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Re: Given line L, and a parallel line that runs through point  [#permalink]

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New post 13 Jun 2015, 05:02
1 - We need to calculate the second equation line.
Since it is parallel to the other one, slope is 3/4[x] and C is Y = 3/4[X] + C i.e. 5 = 3/4[-1] + C
C = 23/4
Y = 3/4[X] + 23/4

2 - Then calculate the distance between two parallel lines
|b-c|/ [M²+1]sqrt(2)
|-1/2 - 23/4| / [3/4 + 1]sqrt(2) = 5
5 is the perpendicular distance between the two lines

3 - Calculate the perimeter
2x[5+9] = 28
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Re: Given line L, and a parallel line that runs through point  [#permalink]

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New post 03 Nov 2017, 09:37
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Given line L, and a parallel line that runs through point &nbs [#permalink] 03 Nov 2017, 09:37
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