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

Question

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}

Skywalker18 · Accepted Answer

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  	riangle  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

robu · Answer

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!

Ekland · Answer

The explanation above seems like magic.
Can someone offer a learnable explanation?

Skywalker18 · Answer

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 riangle which has vertices at (0,0) , (4,0) and (0,4) We have an isosceles right riangle => \angle bca =45 degrees Coordinates of point a are (3,0) and point c are (4,0) Hope it helps!!

PMZ21 · Answer

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/ (2)^1/2
==> Option D

sjavvadi · Answer

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.

Shruti0805 · Answer

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.

atomicmass · Answer

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}

Usernamevisible · Answer

First check what the two lines really are.

Given:

x + y = 3

and

2x + 2y = 8

Simplify the second equation by dividing by 2:

x + y = 4

So the lines are:

x + y = 3

x + y = 4

They have the same slope, so they are parallel.

For two parallel lines:

ax + by + c1 = 0

and

ax + by + c2 = 0

distance formula:

|c2 − c1| / √(a^2 + b ^ 2)

Rewrite both lines:

x + y − 3 = 0

x + y − 4 = 0

Thus:

a = 1, b = 1, c1 = −3, c2 = −4

Distance:

|−4 − (−3)| / √(1^2 + 1^2)

= 1 / √2

Rationalize:

√2 / 2

So the shortest distance is:

√2 / 2

That matches choice D.

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

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GMAT Problem Solving (PS) Questions

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

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