2.1 Equations of Lines

Slope of a Line (p.68) The slope or “steepness” of a line containing the points is the ratio of the change in y over the change in x. The “change in y” or The “change in x” or The slope, m =

Point-Slope Form of the Equation of a Line (p.69)

For a line with slope m and passing through the point (x1, y1), the Point-Slope Form of the equation of the line is y = m(x  x1) + y1 or

y  y1 = m(x  x1)

Example 1 (p.69): Determine the point-slope form for the equation of the line passing through the points (–2, –3) and (1, 3). Then plot the points and graph the line by hand. Solution: Calculate the slope: m = = 2

Now, substitute the coordinates of either point for (x1, y1) and 2 for m

Substituting (1, 3) for (x1, y1) and 2 for m

y = m(x –x1) + y1 y = 2(x – 1) + 3 This is the Point–Slope Form Continue solving for y, y = 2x – 2 + 3 y = 2x + 1

Substituting (–2, –3) for (x1, y1) and 2 for m

y = m(x –x1) + y1 y = 2(x – (–2)) + (–3) y = 2(x + 2) – 3 This is the Point–Slope Form

Continue solving for y, y = 2x + 4 – 3 y = 2x + 1 This is the same as for (1, 3).

Here’s the graph: Slope-Intercept Form of the Equation of a Line (p.70) For a line with slope m and y-intercept b, the Slope-Intercept Form of the equation of the line is y = mx + b

Example 3 (p.70): Find the slope-intercept form for the equation of the line passing through the points (–2, 1) and (2, 3). Solution: Calculate the slope: m =

Now use the coordinates of either point to find b. Using (2, 3), substitute 2 for x and 3 for y.

The equation is

Standard Form of the equation of a line (p.71) is the Standard Form for the equation of a line. a, b and c are constants with a and b not both zero. Examples of linear equations written in Standard Form:

Finding Intercepts (p.72) The x-intercept is the point where the line intersects the x-axis. The coordinates of any point on the x-axis are (#,0). To find the x-intercept, let y = 0 in the equation and solve for x.

The y-intercept is the point where the line intersects the y-axis. The coordinates of any point on the y-axis are (0,#). To find the y-intercept, let x = 0 in the equation and solve for y.

Example 5 (p.72): Find the x- and y-intercepts for the line whose equation is 4x + 3y = 6. Use the intercepts to graph the equation. Solution: To find the x-intercept, let y = 0 and solve for x.

The x-intercept is (1.5, 0)

To find the y-intercept, let x = 0 and solve for y.

The y-intercept is (0, 2)

Here’s the graph:

Horizontal Lines Graph of a constant function f Formula: f (x) = b Horizontal line has slop= 0 and y-intercept = b.

Note that regardless of the value of x, the value of y is always 3.

Vertical Lines Cannot be represented by a function Slope is undefined Equation is: x = k

Note that regardless of the value of y, the value of x is always 3.

Equation is x = 3 or x + 0y = 3 Parallel Lines (p.73)

Two distinct lines with slopes m1 and m2 and y-intercepts , neither of which is vertical, are parallel if and only if they have the same slope, m1 = m2, and different y-intercepts,

NOTE: The phrase “if and only if” (abbreviated “iff”) is used when two statements are mathematically equivalent.

1. If two distinct lines are parallel, then m1 = m2 and 2. If two distinct lines have equal slopes and different y-intercepts, then they are parallel.

Perpendicular Lines (p.74)

Two lines with nonzero slopes m1 and m2 are perpendicular if and only if the product of their slopes is –1. m1m2 = –1

This implies that for perpendicular lines, m1 and m2 are negative reciprocals. Example 8 (p.74): Find the slope-intercept form of the line perpendicular to and passing through the point (– 2, 1). Graph the lines.

Solution: The line has slope The negative reciprocal is Use the slope-intercept form of a line.

The equation is

Here’s the graph:

Modeling Data: Interpolation and Extrapolation (pp.75 and 76) Interpolation occurs when we estimate values between given data points. (p.75) Extrapolation occurs when we estimate values outside the given data. (p.76)

Interpolation is usually more accurate than extrapolation. (p.76)

Example 10 (p.76): Modeling Investments for Cloud Computing The table lists the investments in billions of dollars for cloud computing for selected years.

Year 2005 2006 2007 2008 2009 Investments 26 113 195 299 374 a) Make a scatterplot of the data.

b) Find a linear function in point-slope form. First data point (2005, 26) Last data point (2009, 374) Find the slope.

Equation is c) Graph the data and y = f(x) in the same xy-plane. d) Interpret the slope of the graph of y = f(x).

The positive slope of 87 indicates that investments increased an average of $87 billion per year from 2005 to 2009. e) Estimate the investment in cloud computing in 2014. Does your answer involve interpolation or extrapolation?

To estimate the investments in 2014, we can evaluate for x = 2014. y = 87(2014-2005) + 26 = 809 This model predicts an $809 billion investment in cloud computing during 2014. This result involves extrapolation because 2014 is “outside” of the interval from 2005 to 2009.