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Straight-Line Equations: Parallel and
     Perpendicular Lines (page 3 of 3)

Sections: Slope-intercept form, Point-slope form, Parallel and perpendicular lines

There is one other consideration for straight-line equations: finding parallel and perpendicular lines. Here is the usual format for the question:

• Given the line 2x – 3y = 9 and the point (4, –1), find lines through the point that
  are (a) parallel to the given line and (b) perpendicular to it.

In other words, they've given me a reference line -- 2x – 3y = 9 -- that I'll be comparing to, and some point somewhere else on the plane -- namely, (4, –1). Then they want me to find the line through (4, –1) that is parallel to (that has the same slope as) 2x – 3y = 9. On top of that, they then want me to find the line through (4, –1) that is perpendicular to (that has a slope that is the negative reciprocal of the slope of) 2x – 3y = 9.

Clearly, the first thing I need to do is solve "2x – 3y = 9" for "y=", so that I can find my reference slope: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

2x – 3y = 9
      –3y = –2x + 9
          y = ( ²/₃)x – 3

So the reference slope from the reference line is m = ²/₃.

Since a parallel line has an identical slope, then the parallel line through (4, –1) will have slope m = ²/₃. Hey, now I have a point and a slope! So I'll use the point-slope form to find the line:

y – (–1) = ( ²/₃)(x – 4)
y + 1 = ( ²/₃)x – ⁸/₃
y = ( ²/₃)x – ⁸/₃ – ³/₃

y = ( ²/₃)x – ¹¹/₃

This is the parallel line that they asked for.

For the perpendicular line, I have to find the perpendicular slope. The reference slope is m = ²/₃, and, for the perpendicular slope, I'll flip this slope and change the sign. Then the perpendicular slope is m = – ³/₂. So now I can do the point-slope form. Note that the only change from the calculations I just did is that the slope is different now.

y – (–1) = (– ³/₂)(x – 4)
y + 1 = (– ³/₂)x + 6

y = (– ³/₂)x + 5

If a question asks you whether two given lines are "parallel, perpendicular, or neither", you must answer that question by finding their slopes, not by drawing a picture! Pictures can only give you a rough idea of what is going on, but you cannot tell "by looking" that lines with slopes of, say, m₁ = 1.00 and m₂ = 0.99 are NOT parallel, because they'll sure look parallel on their graphs. But since 1.00 does not equal 0.99, the lines are not parallel. Find the slopes; don't just draw the pictures.

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