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Section 2.4: Equations of Lines

Mr. Dreyer

Algebra 2 Lv 1

1Warmup

Welcome/expectations signatures on desk
Got calculators?
Write down homework: Read Chap 2.3 (p. 82-94). Problems (p. 95-98): 1-12, 13-57 odd, 59, 62, 66, 69, 70, 75, 79, 83, 89, 91
How are graphs of functions and graphs of equations different? How are they similar?

2Quiz soon

Not sure exactly when, but be ready...

3Slope-intercept again

y = m x + b

For example, if $m = 2$ and $b = - 1$ , our equation is

y = 2 x + (- 1)

Reminders:

$m$ is the slope
$b$ is the $y$ -intercept
If we know the slope and $y$ -intercept we know the equation
If we know the equation (in this form) we know the slope and $y$ -intercept, and therefore the graph

4Point-slope form

But what if we just know the slope and some other point, $(x_{0}, y_{0})$ , on the graph, and we don't know the $y$ -intercept? Then we can write the equation in the point-slope form:

y - y_{0} = m (x - x_{0})

How do we know that this equation give the proper graph? Since two points determine a line we just have to show that two solutions to the equation are on the graph.

For our first point, let's pick the original point $(x_{0}, y_{0})$ . Is it a solution to the equation (does it make the equation true)? Let's plug it in:

\begin{array}{rcl} y - y_{0} & = & m (x - x_{0}) \\ y_{0} - y_{0} & = & m {(x_{0} - x_{0})}_{} \\ 0 & = & 0 \end{array}

What other point is on the graph? Well, if we add $1$ to $x$ we should add $m$ to $y$ (why?)

\begin{array}{rcl} y - y_{0} & = & m (x - x_{0}) \\ (y_{0} + m) - y_{0} & = & m {((x_{0} + 1) - x_{0})}_{} \\ m & = & m \end{array}

So that shows that this equation does indeed produce the line with slope $m$ passing through the point $(x_{0}, y_{0})$ .

(show graphical example here)

If we know the point-slope form we can convert to slope-intercept to find the $y$ -intercept:

\begin{array}{rcl} y - y_{0} & = & m (x - x_{0}) \\ y_{} & = & m {(x_{} - x_{0})}_{} + y_{0} \\ = & m x - m x_{0} + y_{0} \\ = & m x + (- m x_{0} + y_{0}) \end{array}

So, starting with the slope-intercept form we can determine that the $y$ -intercept is

- m x_{0} + y_{0}

5Equations of parallel and perpendicular lines

If you have a linear equation (in slope-intercept, point-slope or whatever) and want an equation of a parallel line through the point $(x_{1}, y_{1})$ , just find the slope $m$ and use the point-slope form:

y - y_{1} = m (x - x_{1})

How about a line perpendicular to the first one? Remember that if a line has a slope of $m$ , perpendiculars have a slope of $- \frac{1}{m}$ .

6Just have two points

If you just have two points $(x_{0}, y_{0})$ and $(x_{1}, y_{1})$ , you can find the slope

\frac{rise}{run} = \frac{y_{1} - y_{0}}{x_{1} - x_{0}}

Then you can use any of the above techniques.

7The real world: CO $_{2}$ in the atmosphere

In 1960 there were about 320 ppm of CO $_{2}$ in the atmosphere. In 2000 there were about 370 ppm. If it keeps growing linearly, in what year will the concentration have doubled (to 640) from 1960 levels?

\begin{array}{rcl} (x_{0}, y_{0}) & = & (1960, 320) \\ (x_{1}, y_{1}) & = & (2000, 370) \\ slope & = & \frac{y_{1} - y_{0}}{x_{1} - x_{0}} \\ = & \frac{370 - 320}{2000 - 1960} \\ = & \frac{5}{4} \end{array}

Then an equation for the line is (using point-slope):

y - 320 = \frac{5}{4} (x - 1960)

Solving for $y = 640$ we get

\begin{array}{rcl} 640 - 320 & = & \frac{5}{4} (x - 1960) \\ 320 & = & \frac{5}{4} x - 2450 \\ 320 + 2450 & = & \frac{5}{4} x \\ 2770 & = & \frac{5}{4} x \\ 2216 & = & x \end{array}

8Direct variation

Back to slope-intercept, there's a special case where the $y$ -intercept is 0. In that case we can write the slope-intercept form as

y = m x

In that case we sometimes rename the slope constant $k$ instead of $m$ and we call it the constant of variation, and we say that $y$ varies directly with $x$ . We may write it in two equivalent forms:

\begin{array}{rcl} y & = & k x \\ \frac{y}{x} & = & k \end{array}