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# Distance b/w 2 lines

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Manager
Joined: 15 Feb 2011
Posts: 162

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31 Jul 2011, 21:20
What is the shortest distnace b/w the following 2 lines:
x+y=3 and
2x+2y=8

Well this seemed to be an easy one on the onset but i got it wrong..um sure i goofed up somewhere..

pls help me with the answer explanation. I will post the answer later.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10780
Location: Pune, India
Re: Distance b/w 2 lines  [#permalink]

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01 Aug 2011, 02:12
3
DeeptiM wrote:
What is the shortest distnace b/w the following 2 lines:
x+y=3 and
2x+2y=8

Well this seemed to be an easy one on the onset but i got it wrong..um sure i goofed up somewhere..

pls help me with the answer explanation. I will post the answer later.

The shortest distance between any two lines on the xy axis would be either 0 (the lines will intersect somewhere) or the lines will be parallel.

The two lines are:
x + y = 3
x + y = 4
They are parallel.
(Recall that the test of parallel lines is a1/a2 = b1/b2)
They intersect the x axis at x = 3 and x = 4 and at y axis at y = 3 and y = 4.

Now there are many ways of getting the distance between them.
The first that comes to my mind is using the little triangle abc. ab is dropped perpendicular to the line.
Co-ordinates of a are (3, 0) and of c are (4, 0).
Attachment:

Ques3.jpg [ 6.25 KiB | Viewed 6993 times ]

In the little triangle, ac = 1 unit, angle bca is 45 degrees so it is a 45-45-90 triangle. Since ac = 1 unit, $$ab = 1/\sqrt{2}$$
Recall that the sides in a 45-45-90 are 1:1:$$\sqrt{2}$$

The distance between the 2 lines = ab = $$1/\sqrt{2}$$ or $$\sqrt{2}/2$$
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Karishma
Veritas Prep GMAT Instructor

Manager
Joined: 15 Feb 2011
Posts: 162
Re: Distance b/w 2 lines  [#permalink]

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01 Aug 2011, 04:06
VeritasPrepKarishma wrote:
DeeptiM wrote:
What is the shortest distnace b/w the following 2 lines:
x+y=3 and
2x+2y=8

Well this seemed to be an easy one on the onset but i got it wrong..um sure i goofed up somewhere..

pls help me with the answer explanation. I will post the answer later.

The shortest distance between any two lines on the xy axis would be either 0 (the lines will intersect somewhere) or the lines will be parallel.

The two lines are:
x + y = 3
x + y = 4
They are parallel.
(Recall that the test of parallel lines is a1/a2 = b1/b2)
They intersect the x axis at x = 3 and x = 4 and at y axis at y = 3 and y = 4.

Now there are many ways of getting the distance between them.
The first that comes to my mind is using the little triangle abc. ab is dropped perpendicular to the line.
Co-ordinates of a are (3, 0) and of c are (4, 0).
Attachment:
Ques3.jpg

In the little triangle, ac = 1 unit, angle bca is 45 degrees so it is a 45-45-90 triangle. Since ac = 1 unit, $$ab = 1/\sqrt{2}$$
Recall that the sides in a 45-45-90 are 1:1:$$\sqrt{2}$$

The distance between the 2 lines = ab = $$1/\sqrt{2}$$ or $$\sqrt{2}/2$$

Thanks Karishma but how did we know angle bca is 45 degrees..
I understood your point of 45-45-90 triangle but how did we arrive at sqrt2/2
Pls help..
Manager
Joined: 27 Apr 2008
Posts: 151
Re: Distance b/w 2 lines  [#permalink]

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02 Aug 2011, 16:06
Since you know that triangle BCA is a right angle isosceles triangle, you know that bc=ba. Use the
Pythagorean Theorem to solve for bc.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10780
Location: Pune, India
Re: Distance b/w 2 lines  [#permalink]

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02 Aug 2011, 20:27
3
DeeptiM wrote:

Thanks Karishma but how did we know angle bca is 45 degrees..
I understood your point of 45-45-90 triangle but how did we arrive at sqrt2/2
Pls help..

When you draw the line, you notice that its x and y intercept is the same.
i.e. x + y = 3 intersects x axis at 3 and y axis also at 3. So it forms an isosceles triangle.
Similarly, x + y = 4 intersects x axis at 4 and y axis also at 4.
It also forms an isosceles triangle so angle bca is 45 degrees.
Since angle abc is 90 (we dropped a perpendicular), angle bac will also be 45 degrees (to make the sum 180). So ab = ac.
In an isosceles right triangle, the ratio of the sides is $$1:1:\sqrt{2}$$ where $$\sqrt{2}$$ is the hypotenuse.
Since we know that the hypotenuse is actually 1, the measure of equal sides will be 1/$$\sqrt{2}$$. Multiply and divide this by $$\sqrt{2}$$ to get $$\sqrt{2}$$/2.

Check how to deal with ratios here:
http://www.veritasprep.com/blog/2011/03 ... of-ratios/
_________________
Karishma
Veritas Prep GMAT Instructor

Retired Moderator
Joined: 20 Dec 2010
Posts: 1462
Re: Distance b/w 2 lines  [#permalink]

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02 Aug 2011, 23:27
4
DeeptiM wrote:
What is the shortest distnace b/w the following 2 lines:
x+y=3 and
2x+2y=8

Well this seemed to be an easy one on the onset but i got it wrong..um sure i goofed up somewhere..

pls help me with the answer explanation. I will post the answer later.

Distance between two parallel lines $$y=mx+b$$ and $$y=mx+c$$ can be found by the formula:
$$D=\frac{|b-c|}{\sqrt{m^2+1}}$$

Here:
$$x+y=3$$
||
$$y=-x+3$$--------------------1

$$2x+2y=8$$
||
$$y=-x+4$$---------------------2

$$D=\frac{|b-c|}{\sqrt{m^2+1}}$$

$$b=3$$
$$c=4$$
$$m=-1$$

$$D=\frac{|3-4|}{\sqrt{(-1)^2+1}}$$

$$D=\frac{1}{\sqrt{2}}$$

Formula extracted from:
math-coordinate-geometry-87652.html
Manager
Joined: 15 Feb 2011
Posts: 162
Re: Distance b/w 2 lines  [#permalink]

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03 Aug 2011, 00:26
VeritasPrepKarishma wrote:
DeeptiM wrote:

Thanks Karishma but how did we know angle bca is 45 degrees..
I understood your point of 45-45-90 triangle but how did we arrive at sqrt2/2
Pls help..

When you draw the line, you notice that its x and y intercept is the same.
i.e. x + y = 3 intersects x axis at 3 and y axis also at 3. So it forms an isosceles triangle.
Similarly, x + y = 4 intersects x axis at 4 and y axis also at 4.
It also forms an isosceles triangle so angle bca is 45 degrees.
Since angle abc is 90 (we dropped a perpendicular), angle bac will also be 45 degrees (to make the sum 180). So ab = ac.
In an isosceles right triangle, the ratio of the sides is $$1:1:\sqrt{2}$$ where $$\sqrt{2}$$ is the hypotenuse.
Since we know that the hypotenuse is actually 1, the measure of equal sides will be 1/$$\sqrt{2}$$. Multiply and divide this by $$\sqrt{2}$$ to get $$\sqrt{2}$$/2.

Check how to deal with ratios here:
http://www.veritasprep.com/blog/2011/03 ... of-ratios/

Thanks for being patient with me Karishma..+1 for you
Manager
Joined: 03 Oct 2013
Posts: 76
Re: Distance b/w 2 lines  [#permalink]

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13 Nov 2016, 23:29
The two lines are parallel lines. Once we know that, we can use the distance between parallel lines formula. Or if you prefer drawing figures, easier to draw a line from origin perpendicular to both these lines and calculate the distance.
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Posts: 15592
Re: Distance b/w 2 lines  [#permalink]

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24 Feb 2019, 06:54
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Re: Distance b/w 2 lines   [#permalink] 24 Feb 2019, 06:54