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

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Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139569 [0], given: 12794

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

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15 Dec 2017, 01:09
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45% (medium)

Question Stats:

46% (01:22) correct 54% (00:43) wrong based on 38 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

[Reveal] Spoiler:
Attachment:

2017-12-15_1259.png [ 4.85 KiB | Viewed 382 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 139569 [0], given: 12794

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Joined: 22 May 2016
Posts: 1253

Kudos [?]: 464 [0], given: 683

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

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15 Dec 2017, 12:22
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

[Reveal] Spoiler:
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 $$\frac{\sqrt2}{\sqrt2}$$

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

[Reveal] Spoiler:
C

Attachment:

linesmandn.png [ 18.05 KiB | Viewed 211 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 $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

[Reveal] Spoiler:
C

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Kudos [?]: 464 [0], given: 683

In the rectangular coordinate system above, if the equation of m is y   [#permalink] 15 Dec 2017, 12:22
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