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What is the shortest distance between the following 2 lines: x + y = 3

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What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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What is the shortest distance between the following 2 lines: x + y = 3 and 2x + 2y = 8?

(A) 0
(B) 1/4
(C) 1/2
(D) \(\frac{{\sqrt 2}}{2}\)
(E) \(\frac{{\sqrt 2}}{4}\)
[Reveal] Spoiler: OA

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 12 Feb 2016, 03:03
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The shortest distance between any two lines on the xy axis would be either
0 - the lines intersect
or the lines are parallel.

The two lines are:
x + y = 3
2x + 2y = 8
=>x + y = 4
The given 2 lines are parallel. (a1/a2 = b1/b2 )

The lines intersect x and y axis - (3,0) , (4,0) , (0,3) and (0,4)


In \(\triangle\) ABC
ac= 1 ,
\(\angle\) bca =45 degrees
so it is a 45-45-90 triangle. Since ac = 1 ,
Sin 45 = ab/ac = ab/1 =
=1/2
=> ab =1/(2)^(1/2)
The sides in a 45-45-90 are 1:1:2√2

The distance between the 2 lines = ab = 1/(2)^(1/2) or (2)^(1/2) / 2

Answer D
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GEO.PNG [ 6.25 KiB | Viewed 2742 times ]


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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 15 Feb 2016, 12:40
The explanation above seems like magic.
Can someone offer a learnable explanation?

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 15 Feb 2016, 20:26
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Nez wrote:
The explanation above seems like magic.
Can someone offer a learnable explanation?


Hi Nez ,
I did not intend to give a magical explanation :) . Let me specify the steps that were followed -
The steps that i have followed are -
1. The given lines are parallel
2. Find where the given 2 lines intersect with X and Y axis .
The line x+y=3 (-- line 1) will pass through (3,0) and (0,3)
and the line x + y = 4 (-- line 2) will pass through (4,0) and (0,4)
3. The shortest distance between 2 parallel lines is the perpendicular distance.
So , a perpendicular- ab has been dropped from line 2 to line 1.


In the \(\triangle\) which has vertices at (0,0) , (4,0) and (0,4)
We have an isosceles right \(\triangle\)
=>\(\angle\) bca =45 degrees

Coordinates of point a are (3,0) and point c are (4,0)

Hope it helps!! :)

Skywalker18 wrote:
The shortest distance between any two lines on the xy axis would be either
0 - the lines intersect
or the lines are parallel.

The two lines are:
x + y = 3 -- line 1
2x + 2y = 8
=>x + y = 4 -- line 2
The given 2 lines are parallel. (a1/a2 = b1/b2 )

The lines intersect x and y axis - (3,0) , (4,0) , (0,3) and (0,4)


In \(\triangle\) ABC
ac= 1 ,
\(\angle\) bca =45 degrees
so it is a 45-45-90 triangle. Since ac = 1 ,
Sin 45 = ab/ac = ab/1 =
=1/2
=> ab =1/(2)^(1/2)
The sides in a 45-45-90 are 1:1:2√2

The distance between the 2 lines = ab = 1/(2)^(1/2) or (2)^(1/2) / 2

Answer D

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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Bunuel wrote:
What is the shortest distance between the following 2 lines: x + y = 3 and 2x + 2y = 8?

(A) 0
(B) 1/4
(C) 1/2
(D) \(\frac{{\sqrt 2}}{2}\)
(E) \(\frac{{\sqrt 2}}{4}\)


first write the equations in the form:

x+y-3=0
x+y-4=0

Now the formula for shortest/ perpendicular distance = Mod(C1-C2)/sqrt (a^2+b^2)
i.e.
Mode (C1-C2) = mode (-3-(-4)) =1
Sqrt(a^2+b^2)= 1^2+1^2 =root 2.

answer= 1/sqrt2= sqrt2/2.

I hope it makes some sense!

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 31 Mar 2016, 14:12
The explanation by Skywalker18 though with good intention. is a little bit unfortunate in my opinion.
Sin and cosine and cotangent are not part GMAT scope.
Are they?
Well why would you introduce something remarkably off into this when you are addressing not math graduates but GMAT students.
If I know cosines, it's in my pocket.
There are far easier methods than the silly cosines.
I thought we are helping ourselves here not showing off our mathry.

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What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 26 Feb 2017, 06:49
robu wrote:
Bunuel wrote:
What is the shortest distance between the following 2 lines: x + y = 3 and 2x + 2y = 8?

(A) 0
(B) 1/4
(C) 1/2
(D) \(\frac{{\sqrt 2}}{2}\)
(E) \(\frac{{\sqrt 2}}{4}\)


first write the equations in the form:

x+y-3=0
x+y-4=0

Now the formula for shortest/ perpendicular distance = Mod(C1-C2)/sqrt (a^2+b^2)
i.e.
Mode (C1-C2) = mode (-3-(-4)) =1
Sqrt(a^2+b^2)= 1^2+1^2 =root 2.

answer= 1/sqrt2= sqrt2/2.

I hope it makes some sense!



Hi robu,

In the formula that you've used, can we take the a and b values from either of the 2 equations ?
In the current qsn, the values are equal, hence clarifying.

Also, can this formula be used even without knowing that the lines are parallel ?

Thanks in advance! :)

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What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 10 May 2017, 12:03
Skywalker18 wrote:
Nez wrote:
The explanation above seems like magic.
Can someone offer a learnable explanation?


Hi Nez ,
I did not intend to give a magical explanation :) . Let me specify the steps that were followed -
The steps that i have followed are -
1. The given lines are parallel
2. Find where the given 2 lines intersect with X and Y axis .
The line x+y=3 (-- line 1) will pass through (3,0) and (0,3)
and the line x + y = 4 (-- line 2) will pass through (4,0) and (0,4)
3. The shortest distance between 2 parallel lines is the perpendicular distance.
So , a perpendicular- ab has been dropped from line 2 to line 1.


In the \(\triangle\) which has vertices at (0,0) , (4,0) and (0,4)
We have an isosceles right \(\triangle\)
=>\(\angle\) bca =45 degrees

Coordinates of point a are (3,0) and point c are (4,0)

Hope it helps!! :)

Skywalker18 wrote:
The shortest distance between any two lines on the xy axis would be either
0 - the lines intersect
or the lines are parallel.

The two lines are:
x + y = 3 -- line 1
2x + 2y = 8
=>x + y = 4 -- line 2
The given 2 lines are parallel. (a1/a2 = b1/b2 )

The lines intersect x and y axis - (3,0) , (4,0) , (0,3) and (0,4)


In \(\triangle\) ABC
ac= 1 ,
\(\angle\) bca =45 degrees
so it is a 45-45-90 triangle. Since ac = 1 ,
Sin 45 = ab/ac = ab/1 =
=1/2
=> ab =1/(2)^(1/2)
The sides in a 45-45-90 are 1:1:2√2

The distance between the 2 lines = ab = 1/(2)^(1/2) or (2)^(1/2) / 2

Answer D


Excellent Explanation, Skywalker. The key is to understand that the triangle is Isosceles and angles are 45. That is super GMAT question and answer.
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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 13 May 2017, 10:14
can this question be solved using formula :Distance between two parallel lines y=mx+ c and y=mx+c is
D= \(\frac{|b-c|}{\sqrt{(m^2+1)}}\)

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 23 May 2017, 09:08
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Two lines:

L1: x + y = 3 --> y = (-1)x + 3, where slope m1 = -1 and b = 3
L2: 2x + 2y = 8 --> 2y = -2x + 8 --> y = (-1) + 4, where slope m2 = -1 and c = 4

Since the two slopes are equal, line 1 is parallel to line 2.

Shorted distance between two parallel lines can be found by using the formula: | b-c | / (m^2 + 1)^1/2
==> |4-3| / [(-1)^2) + 1]^1/2
==> 1/ (2)^1/2
==> Option D

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 23 May 2017, 10:22
I like the 45-45-90 solution because it explains the solution conceptually.

Just in case as an alternate method I have attached the solution using the the "mid-point" information and then the distance between two points.
Attachments

Untitled.png
Untitled.png [ 79.21 KiB | Viewed 1430 times ]


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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 23 May 2017, 10:29
I just realized that the figure I have used is wrong. It needs to be a mirror image i.e. the points of intersection on x-axis are (3,0) and (4,0). the concept is the same though and the answer remains the same (due to the squaring in the distance formula).
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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 23 May 2017, 10:31
Bunuel wrote:
What is the shortest distance between the following 2 lines: x + y = 3 and 2x + 2y = 8?

(A) 0
(B) 1/4
(C) 1/2
(D) \(\frac{{\sqrt 2}}{2}\)
(E) \(\frac{{\sqrt 2}}{4}\)



x + y = 3
y= -x + 3 (Y=mx +b format)
and 2x + 2y = 8
y = -x + 4

Shortest distance between two parallel lines (m1 = m2) = \((|b-c|)/\sqrt{m^2 +1}\)
on solving it is \(1/\sqrt{2}\) hence option D
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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 23 May 2017, 10:43
sjavvadi wrote:
I like the 45-45-90 solution because it explains the solution conceptually.

Just in case as an alternate method I have attached the solution using the the "mid-point" information and then the distance between two points.



How can we say that the shortest distance will be the line joining the mid points?

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 23 May 2017, 23:05
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Bunuel wrote:
What is the shortest distance between the following 2 lines: x + y = 3 and 2x + 2y = 8?

(A) 0
(B) 1/4
(C) 1/2
(D) \(\frac{{\sqrt 2}}{2}\)
(E) \(\frac{{\sqrt 2}}{4}\)



Lets consider the points where Line 1 and Line 2 intersect the X and Y axes:

L1 : x + y = 3
=> y= -x + 3
Substitue x=0 to get y=3 and y=0 to get x=3.
So, the 2 points are (0,3) and (3,0).

Similarly for L2 : 2x+2y = 8
=> y = -x + 4
Points are (0,4) and (4,0).


Check, in the attached figure, both the lines are parallel to each other and create a right triangle with the X and Y axes.
Now, in right triangles, we know that Base*Height = Hypotenuse*Perpendicular dropped from the 90 degrees vertex.
Also notice because the 2 shorter sides are equal, that makes it a 45-45-90 triangle, thereby, the sides would be in the ratio \(1:1:\sqrt{2}\)

So, for the 1st triangle, using the distance formula for both the smaller sides, \(3*3 = 3\sqrt{2} * X\)
=> \(X = \frac{3}{\sqrt{2}}\)

Similarly, for the 2nd triangle, \(4*4 = 4\sqrt{2} * Y\)
=> Y = \(\frac{4}{\sqrt{2}}\)

Shortest distance is always the perpendicular distance between 2 lines, if they're parallel, so in our case, the distance should be (Y-X)
= \(\frac{1}{\sqrt{2}}\)
Option D.
Attachments

IMG_20170524_113119.jpg
IMG_20170524_113119.jpg [ 3.42 MiB | Viewed 1375 times ]

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 24 May 2017, 10:24
nishantt7 wrote:
sjavvadi wrote:
I like the 45-45-90 solution because it explains the solution conceptually.

Just in case as an alternate method I have attached the solution using the the "mid-point" information and then the distance between two points.



How can we say that the shortest distance will be the line joining the mid points?



Need not be the mid-points. Basically, it is the distance between the lines. So we need to find the perpendicular distance between the two lines. since these are parallel lines with the right angle at (0,0) I picked the mid points.
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What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 24 May 2017, 18:16
sjavvadi wrote:
nishantt7 wrote:
sjavvadi wrote:
I like the 45-45-90 solution because it explains the solution conceptually.

Just in case as an alternate method I have attached the solution using the the "mid-point" information and then the distance between two points.



How can we say that the shortest distance will be the line joining the mid points?



Need not be the mid-points. Basically, it is the distance between the lines. So we need to find the perpendicular distance between the two lines. since these are parallel lines with the right angle at (0,0) I picked the mid points.


I don't think there's a need to pick a point. Since the shape is fairly simple I think we don't have to get the numbers involved.

We know 2 lines are parallel and both create a 45 degree angle with the axis.
Also, if you shift the entire x + y = 4 to the left by 1 unit, you will get x + y = 3.

Therefore 1 = length of the bigger side of the isosceles right triangle created by drawing a random perpendicular line between the 2. The smaller side = 1/sqrt2.

It's a lot of words but it's actually much faster when you visualize this in your head. Took me 12s with no pen using this method. Hope this helps.

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 16 Jun 2017, 06:24
Bunuel wrote:
What is the shortest distance between the following 2 lines: x + y = 3 and 2x + 2y = 8?

(A) 0
(B) 1/4
(C) 1/2
(D) \(\frac{{\sqrt 2}}{2}\)
(E) \(\frac{{\sqrt 2}}{4}\)


The most simple method is indeed to construct a triangle- this is more logical than applying a formula simply because we know the base of triangle as the distance between (4,0) and (3,0) = 1 so according to the ratio of an isosceles right triangle

1: 1: \sqrt{2}

You simply work backwards-

1/\sqrt{2} is the length of the side however, you need to rationalize the denominator which is the unexplained step in many of the solutions here

1/\sqrt{2} * \sqrt{2}/\sqrt{2}=

\sqrt{2}/2

Thus
"D"

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 16 Jun 2017, 16:15
coolkl wrote:
Skywalker18 wrote:
Nez wrote:
The explanation above seems like magic.
Can someone offer a learnable explanation?


Hi Nez ,
I did not intend to give a magical explanation :) . Let me specify the steps that were followed -
The steps that i have followed are -
1. The given lines are parallel
2. Find where the given 2 lines intersect with X and Y axis .
The line x+y=3 (-- line 1) will pass through (3,0) and (0,3)
and the line x + y = 4 (-- line 2) will pass through (4,0) and (0,4)
3. The shortest distance between 2 parallel lines is the perpendicular distance.
So , a perpendicular- ab has been dropped from line 2 to line 1.


In the \(\triangle\) which has vertices at (0,0) , (4,0) and (0,4)
We have an isosceles right \(\triangle\)
=>\(\angle\) bca =45 degrees

Coordinates of point a are (3,0) and point c are (4,0)

Hope it helps!! :)

Skywalker18 wrote:
The shortest distance between any two lines on the xy axis would be either
0 - the lines intersect
or the lines are parallel.

The two lines are:
x + y = 3 -- line 1
2x + 2y = 8
=>x + y = 4 -- line 2
The given 2 lines are parallel. (a1/a2 = b1/b2 )

The lines intersect x and y axis - (3,0) , (4,0) , (0,3) and (0,4)


In \(\triangle\) ABC
ac= 1 ,
\(\angle\) bca =45 degrees
so it is a 45-45-90 triangle. Since ac = 1 ,
Sin 45 = ab/ac = ab/1 =
=1/2
=> ab =1/(2)^(1/2)
The sides in a 45-45-90 are 1:1:2√2

The distance between the 2 lines = ab = 1/(2)^(1/2) or (2)^(1/2) / 2

Answer D


Excellent Explanation, Skywalker. The key is to understand that the triangle is Isosceles and angles are 45. That is super GMAT question and answer.


What is the exact rule that validates that this triangle is an isoceles? I instinctively came up with that but need to know a more solid reason why.

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Re: What is the shortest distance between the following 2 lines: x + y = 3 [#permalink]

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New post 05 Oct 2017, 08:30
Bunuel wrote:
What is the shortest distance between the following 2 lines: x + y = 3 and 2x + 2y = 8?

(A) 0
(B) 1/4
(C) 1/2
(D) \(\frac{{\sqrt 2}}{2}\)
(E) \(\frac{{\sqrt 2}}{4}\)


Answer D



Solution


X + Y = 3 -------------------------I
X + Y = 4 -------------------------II

Slope of both the above lines is -1. therefore, both the lines will make angle of 45 with the X coordinate and Y coordinate.

Line perpendicular both of the above lines must have slope 1.
Line Y = X has slope 1.

Substitute the Y = X and point of lines passing at 90 degree.
I -> x = 1.5, y = 1.5 ------------------------------III
II -> x = 2, y = 2 -----------------------------------IV

Using distance formula on III and IV we get \(\frac{1}{{\sqrt 2}}\)
or
\(\frac{{\sqrt 2}}{2}\)

Kudos [?]: 10 [0], given: 48

Re: What is the shortest distance between the following 2 lines: x + y = 3   [#permalink] 05 Oct 2017, 08:30
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What is the shortest distance between the following 2 lines: x + y = 3

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