2.1 Equations of Lines
2.1 Equations of Lines
Slope of a Line (p.68)
The slope or “steepness” of a line containing the points x1,y1 and (x2,y2) is the ratio of the change in y over the change in x.
The “change in y” or ∆y=y2-y1 The “change in x” or ∆x=x2-x1
The slope, m = ∆y∆x=y2-y1x2-x1
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 = 3-(-3)1-(-2)=63 = 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 = 3-12-(-2)=24=12
y=12x+b
Now use the coordinates of either point to find b.
Using (2, 3), substitute 2 for x and 3 for y. 3=122+b
3=1+b
2=b
The equation is y=12x+2
Standard Form of the equation of a line (p.71)
ax+by=c 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:
2x-3y=-6 4y=1 x=-3 -3x+y=12
0x+4y=1 x+0y=-3
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.
4x+3∙0=6 4x=6 x=64=32 or 1.5
The x-intercept is (1.5, 0)
To find the y-intercept, let x = 0 and solve for y.
4∙0+3y=6 3y=6 y=2
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 b1 and b2, neither of which is vertical, are parallel if and only if they have the same slope, m1 = m2, and different y-intercepts, b1≠b2
y=32x+4 y=32x+1
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 b1≠b2.
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 y=-23x+2 and passing through the point (– 2, 1). Graph the lines.
Solution: The line y=-23x+2 has slope -23
The negative reciprocal is 32
Use the slope-intercept form of a line. y=mx+b
1=32-2+b
1=-3+b
4=b
The equation is y=32x+4
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 / 2009Investments / 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. m=374-262009-2005=3484=87
Equation is y=87x-2005+26
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.