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In the rectangular coordinate system above, if the equation of m is y 15 Dec 2017, 02:09
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57% (01:52) correct 43% (01:21) wrong based on 213 sessions

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In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

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2017-12-15_1259.png
2017-12-15_1259.png [ 4.85 KiB | Viewed 11299 times ]
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C
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In the rectangular coordinate system above, if the equation of m is y 15 Dec 2017, 13:22
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Bunuel

In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

Show Spoiler
Attachment:
The attachment 2017-12-15_1259.png is no longer available
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With the formula (My diagram does not apply)
I did not use it, but to do so is standard. So I quote the formula from Bunuel HERE

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

Equation for line \(m\), \(y = x\), to match the exact form of \(y = mx + b\) can be written:
\(y =(1)x + 0\) (all lines that pass through the origin have x- and y-intercepts of 0)
\(b = 0\) = y-intercept, \(m = 1\) = slope

Find the equation of line \(n\), \(y = mx + c\)

Parallel lines have identical slopes. Line \(n\) has slope \(m = 1\)
Use the point on line \(n\) from the graph: (1, 0)
Plug its coordinates into slope-intercept form to find \(c\)
\(y = mx + c\)
\((0) = 1(1) + c\)
\(-c = 1\), so \(c = -1\)
Equation for line \(n: y = x - 1\)

Distance between parallel lines

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

\(m = 1\), \(b = 0\), \(c = -1\)

\(D=\frac{|0-(-1)|}{\sqrt{1^2+1}}\)
\(D=\frac{1}{\sqrt{2}}\)
Does not match the answers. Rationalize the denominator; multiply by \(\frac{\sqrt2}{\sqrt2}\)

\(\frac{1}{\sqrt{2}}\) * \(\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}\)

Answer C

Attachment:
linesmandn.png
linesmandn.png [ 18.05 KiB | Viewed 10532 times ]
Without the formula see diagram

The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A

That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB

A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of \(x : x : x\sqrt{2}\)

The hypotenuse corresponds with \(x\sqrt{2} = 1\): OB = 1
Find length of equal sides \(x\) (one of which, AB, is the distance needed):

\(x\sqrt{2}= 1\)

\(x = \frac{1}{\sqrt{2}}\) = AB = shortest distance between the parallel lines

That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by \(\frac{\sqrt2}{\sqrt2}\)
\(\frac{1}{\sqrt{2}}\) * \(\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}\)

Answer C
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In the rectangular coordinate system above, if the equation of m is y 14 Apr 2018, 06:59
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generis
Quote:
Bunuel[/url]"]
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

Show Spoiler
Attachment:
2017-12-15_1259.png

Attachment:
linesmandn.png
Without the formula see diagram

The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A

That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB

A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of \(x : x : x\sqrt{2}\)

The hypotenuse corresponds with \(x\sqrt{2} = 1\): OB = 1
Find length of equal sides \(x\) (one of which, AB, is the distance needed):

\(x\sqrt{2}= 1\)

\(x = \frac{1}{\sqrt{2}}\) = AB = shortest distance between the parallel lines

That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by \(\frac{\sqrt2}{\sqrt2}\)
\(\frac{1}{\sqrt{2}}\) * \(\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}\)

Answer C
Show more


Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?

thanks.
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In the rectangular coordinate system above, if the equation of m is y 14 Apr 2018, 07:59
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generis
Quote:
Bunuel[/url]"]
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

Show Spoiler
Attachment:
2017-12-15_1259.png

Attachment:
linesmandn.png
Without the formula see diagram

The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A

That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB

A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of \(x : x : x\sqrt{2}\)

The hypotenuse corresponds with \(x\sqrt{2} = 1\): OB = 1
Find length of equal sides \(x\) (one of which, AB, is the distance needed):

\(x\sqrt{2}= 1\)

\(x = \frac{1}{\sqrt{2}}\) = AB = shortest distance between the parallel lines

That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by \(\frac{\sqrt2}{\sqrt2}\)
\(\frac{1}{\sqrt{2}}\) * \(\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}\)

Answer
Show Spoiler
C


Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?

thanks.
Show more

Got it!!!! The Slope is 1 so the angle at the origin will be 45* , as for calculating distance we have to drop a perpendicular one angle will be 90 leaving the third angle as 45*.
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In the rectangular coordinate system above, if the equation of m is y 14 Apr 2018, 16:33
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Quote:
Bunuel wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Show Spoiler
Attachment:
2017-12-15_1259.png
Show more
MT1988
MT1988
generis

Without the formula see diagram

The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A

That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB

A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of \(x : x : x\sqrt{2}\)

The hypotenuse corresponds with \(x\sqrt{2} = 1\): OB = 1
Find length of equal sides \(x\) (one of which, AB, is the distance needed):

\(x\sqrt{2}= 1\)

\(x = \frac{1}{\sqrt{2}}\) = AB = shortest distance between the parallel lines

That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by \(\frac{\sqrt2}{\sqrt2}\)
\(\frac{1}{\sqrt{2}}\) * \(\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}\)

Answer C
Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?

thanks.

Got it!!!! The Slope is 1 so the angle at the origin will be 45* , as for calculating distance we have to drop a perpendicular one angle will be 90 leaving the third angle as 45*.
Show more

BOOM. Excellent. :-D
I came prepared with a diagram
to help, but behold, no need.

You have already explained the answer.
Here is a post that discusses
your analysis of line slopes
that create 45° angles.
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In the rectangular coordinate system above, if the equation of m is y 15 Apr 2018, 03:40
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Bunuel

In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

Show Spoiler
Attachment:
The attachment 2017-12-15_1259.png is no longer available
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My method to solve this:-
Attachments

File comment: for line y = x, the angle between x-axis and the line m will be 45 degrees.
The min distance will the perpendicular distance between the two lines.
In the image, by mistake I've written 1/root(2) = 2/root(2), it should be root(2)/2. Please excuse that mistake
IMG_20180415_160317.jpg
IMG_20180415_160317.jpg [ 1.99 MiB | Viewed 9293 times ]

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In the rectangular coordinate system above, if the equation of m is y 14 Nov 2020, 00:19
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The shortest Distance between the 2 Lines is the Perpendicular Distance between a Point on Each Line

Rule: the Line y = x creates a 45 degree Angle with the X-Axis

Further, our 2nd Line M is represented by the Equation: y = x - 1

To find the Perpendicular Distance between the 2 Lines, we can draw a Right Triangle between Line N and Line M


(1st) From Line M to Line N along the X-Axis, we can draw a Leg = 1 Unit

(2nd)Line N passes through the X-Axis at point (1, 0).

the Point on Line M that is Directly Vertical from Point (1 ,0) will have the same X-Coordinate.
on Line Y = X, this Point will be given by (1 , 1)

Draw a Perpendicular Line from Point (1 , 1) on Line M to Point (1 , 0) on Line N on the X-Axis. This Leg = 1 Unit.

(3rd) Since we have an Isosceles Right Triangle (1 of the Angles is 45 degrees and the 2 Legs are Equal), we can determine the Hypotenuse of our created Triangle ------> sqrt(2)


(4th) We can 1st find the Area of this Triangle. Then, we use the Hypotenuse as Base to find the Perpendicular Distance from the Vertex at Point (1 , 0) to a Point on Line M

Area of our created Right Triangle = (1/2) * (1) * (1) = 1/2

Using the Hypotenuse as Base and naming the Perpendicular Distance between the 2 Lines = H

Area of Triangle = (1/2) = (1/2) * (sqrt(2)) * H

H = 2 / (sqrt(2) *2)

Conjugating the DEN by Multiplying the NUM/DEN both by * sqrt(2) ------->

you end up with H = sqrt(2) / 2 = Perpendicular Distance between the 2 Lines

-C-
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In the rectangular coordinate system above, if the equation of m is y 21 Jun 2022, 11:37
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Bunuel

In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

Show Spoiler
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Two ways.

First, ballpark. This takes maybe 10-15 seconds and has effectively no chance of making a silly mistake. It is, in my opinion, the BEST way to do this question.
Look at the attachment Method 1. Just look at it for a second before you rush off to do math.
Is the distance between the lines m and n more than 1, equal to 1, or less than 1? Less than. A and B are out.
The little red segment is length 1/2. Is the distance between the lines more than 1/2, equal to 1/2, or less than 1/2? More than. D and E are out.
Answer choice C.

Second, geometry.
Look at the attachment Method 2.
The red line makes a 45-45-90 triangle with the x-axis and line n. The hypotenuse is 1, so the legs are \(\frac{1}{\sqrt{2}}\). We need to rationalize the denominator by multiplying by \(\frac{\sqrt{2}}{\sqrt{2}}\). \(\frac{\sqrt{2}}{2}\).
Answer choice C.


ThatDudeKnowsBallparking
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In the rectangular coordinate system above, if the equation of m is y 21 Jun 2022, 22:06
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Bunuel

In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

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Attachment:
The attachment 2017-12-15_1259.png is no longer available
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Attachment:
Screenshot 2022-06-22 at 10.26.49 AM.png
Screenshot 2022-06-22 at 10.26.49 AM.png [ 14.07 KiB | Viewed 4294 times ]

Since equation of line m is y = x, when x = 1, y = 1. So we get an isosceles right triangle (yellow) with both legs of length 1 each. Since it is an isosceles triangle, its sides are in the ratio 1:1:sqrt2. So the hypotenuse length is sqrt2.

Area of this yellow triangle = (1/2) * 1 * 1 = (1/2) * sqrt2 * Altitude (shown by the green line)

Altitude = sqrt2/2

Since altitude is the perpendicular distance, it is the shortest distance between m and n.

Answer (C)
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