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

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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 HEREDistance 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 [ 18.05 KiB | Viewed 1048 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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