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# In the rectangular coordinate system above, if the equation of m is y

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Joined: 02 Sep 2009
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In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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15 Dec 2017, 02:09
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Difficulty:

55% (hard)

Question Stats:

52% (01:24) correct 48% (00:35) wrong based on 119 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

Attachment:

2017-12-15_1259.png [ 4.85 KiB | Viewed 1084 times ]

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In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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15 Dec 2017, 13:22
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1
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

Attachment:
The attachment 2017-12-15_1259.png is no longer available

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 $$\frac{\sqrt2}{\sqrt2}$$

$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

Attachment:

linesmandn.png [ 18.05 KiB | Viewed 868 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 $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

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In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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14 Apr 2018, 06:59
generis 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

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 $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

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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Joined: 20 Dec 2014
Posts: 37
Re: In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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14 Apr 2018, 07:59
1
MT1988 wrote:
generis 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

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 $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

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*.
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Posts: 1833
Re: In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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14 Apr 2018, 16:33
1
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
Attachment:
2017-12-15_1259.png

MT1988 wrote:
MT1988 wrote:
generis wrote:
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 $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

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*.

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

Here is a post that discusses
that create 45° angles.
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that within me there lay an invincible summer.

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Joined: 30 Nov 2016
Posts: 34
GMAT 1: 570 Q47 V21
GMAT 2: 540 Q47 V16
GMAT 3: 560 Q39 V28
In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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15 Apr 2018, 03:40
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

Attachment:
The attachment 2017-12-15_1259.png is no longer available

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 [ 1.99 MiB | Viewed 347 times ]

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In the rectangular coordinate system above, if the equation of m is y &nbs [#permalink] 15 Apr 2018, 03:40
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