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4.1 Linear functions  (Page 12/27)

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− 4 ( 1 4 ) = − 1

Parallel and perpendicular lines

Two lines are parallel lines    if they do not intersect. The slopes of the lines are the same.

f ( x ) = m 1 x + b 1   and   g ( x ) = m 2 x + b 2   are parallel if and only if  m 1 = m 2

If and only if   b 1 = b 2   and   m 1 = m 2 ,   we say the lines coincide. Coincident lines are the same line.

Two lines are perpendicular lines    if they intersect to form a right angle.

f ( x ) = m 1 x + b 1   and   g ( x ) = m 2 x + b 2   are perpendicular if and only if
m 1 m 2 = − 1 , so   m 2 = − 1 m 1

Identifying parallel and perpendicular lines

Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.

f ( x ) = 2 x + 3 h ( x ) = − 2 x + 2 g ( x ) = 1 2 x − 4 j ( x ) = 2 x − 6

Parallel lines have the same slope. Because the functions   f ( x ) = 2 x + 3   and   j ( x ) = 2 x − 6   each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and   1 2   are negative reciprocals, the functions   g ( x ) = 1 2 x − 4   and   h ( x ) = −2 x + 2   represent perpendicular lines.

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Writing the equation of a line parallel or perpendicular to a given line

If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.

Writing equations of parallel lines

Suppose for example, we are given the equation shown.

f ( x ) = 3 x + 1

We know that the slope of the line formed by the function is 3. We also know that the y- intercept is   ( 0 , 1 ) .   Any other line with a slope of 3 will be parallel to   f ( x ) .   So the lines formed by all of the following functions will be parallel to   f ( x ) .

g ( x ) = 3 x + 6 h ( x ) = 3 x + 1 p ( x ) = 3 x + 2 3

Suppose then we want to write the equation of a line that is parallel to   f   and passes through the point   ( 1 , 7 ) .   This type of problem is often described as a point-slope problem because we have a point and a slope. In our example, we know that the slope is 3. We need to determine which value of   b   will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.

y − y 1 = m ( x − x 1 ) y − 7 = 3 ( x − 1 ) y − 7 = 3 x − 3 y = 3 x + 4

So   g ( x ) = 3 x + 4   is parallel to   f ( x ) = 3 x + 1   and passes through the point   ( 1 , 7 ) .

Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.

  1. Find the slope of the function.
  2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.
  3. Simplify.

Finding a line parallel to a given line

Find a line parallel to the graph of   f ( x ) = 3 x + 6   that passes through the point   ( 3 , 0 ) .

The slope of the given line is 3. If we choose the slope-intercept form, we can substitute   m = 3 , x = 3 , and   f ( x ) = 0   into the slope-intercept form to find the y- intercept.

g ( x ) = 3 x + b 0 = 3 ( 3 ) + b b = –9

The line parallel to   f ( x )   that passes through   ( 3 , 0 )   is   g ( x ) = 3 x − 9.

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Writing equations of perpendicular lines

We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the function shown.

Questions & Answers

If c is the cost function for a particular product, find the marginal cost functions and their values at x=10 a. c(x) = 800+ 0.04x + 0.0002x² b. c(x) = 250 + 100x + 0.001x²
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when you reduce an equation to its simplest terms, you can't change the value of the equation. reducing it to y + 5 is equivalent to dividing it by 9 which changes the value. you can multiply it by 1 or 9/9 which would give 9(y + 5). multiplying it by one does not change the value.
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1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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