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

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

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New post 15 Dec 2017, 01:09
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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
2017-12-15_1259.png [ 4.85 KiB | Viewed 382 times ]
[Reveal] Spoiler: OA

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

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New post 15 Dec 2017, 12:22
Bunuel wrote:
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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:
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 \(\frac{\sqrt2}{\sqrt2}\)

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

Answer
[Reveal] Spoiler:
C


Attachment:
linesmandn.png
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 \(\frac{\sqrt2}{\sqrt2}\)
\(\frac{1}{\sqrt{2}}\) * \(\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}\)

Answer
[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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