Linear Equations and Graphing
In math and science we are sometimes given a set of data and need to determine an equation to describe the relationship between them. In this lesson we look at how to find the equation of a line given 2 points.
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One real life situation happened when rangers tried to predict how often the geyser Old Faithful would erupt.
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Your turn: How do you find the equation of the line through points:
$$\color{#9400d3}{(-3,5)(2,1)}$$
Solution:
To solve this problem we will start with a equation for a line. I prefer to use the slope-intercept form: $\color{#9400d3}{y=mx+b}$.
Next we need to find the slope (m) of the line.
$$ \color{#9400d3}{m = \frac{\Delta y}{\Delta x} = \frac{5-1}{-3-2} = \frac{4}{-5}}$$
Now we need to fine the y-intercept. To do that we will pick a point that is on the line, we could use either $(-3,5)$ or $(2,1)$. It does not matter which pair we use, that other pair will be used in checking our equation. Lets use $\color{#9400d3}{(2,1)}$ to determine the y-intercept.
$$
\begin{align*}
y &= \color{#9400d3}{\frac{-4}{5}}x + b\\
\\[1px]
\color{#9400d3}1 &= \frac{-4}{5}*\color{#9400d3}2+b\\
\\[1px]
1 &= \frac{-8}{5}*\color{#9400d3}+b\\
\\[1px]
1 \color{fuchsia}{+\frac{8}{5}} &= \frac{-8}{5}\color{fuchsia}{+\frac{8}{5}}+b\\
\\[1px]
\color{#9400d3}{\frac{13}{5}} &= b\\
\\[1px]
\color{#9400d3}b &= \color{#9400d3}{2\frac{3}{5}}
\end{align*}
$$
Our final step is to add the y-intercept (b) into the formula.
$$ \color{#9400d3}{y = {\frac{-4}{5}}x + 2\frac{3}{5}} $$
Check:
There two ways we can check this equation.
1. by substituting in (-3,5) into the equation and making sure it leads to a true statement (be sure to use the coordinates of the point not used to find the y-intercept.)
$$
\begin{align*}
y &= {\frac{-4}{5}}x + 2\frac{3}{5}\\
\\[1px]
\color{#9400d3}5 &= \frac{-4}{5}*\color{#9400d3}{^-3} + 2\frac{3}{5}\\
\\[1px]
5 &= 2\frac{2}{5}+ 2\frac{3}{5}\\
\\[1px]
5 &= 5
\end{align*}
$$
2.The other way would be to graph the two original points.
Looking at the graph we can see the y-intercept is approximately $2\frac{2}{5}$,
and the slope is $\frac{-4}{5}$.
