Suppose we want to find the equation of a line passing through a known point \(\mathcal{P}(x_0,y_0)\) with a known slope \(m\text{.}\) We need to determine an algebraic equation that we can use to determine if a different point \(\mathcal{Q}(x,y)\) is or is not on the line. If \(\mathcal{Q}\) is on the line, then we can calculate the (known) slope \(m\) from these two points using the slope formula (6.4.1)
\begin{equation*}
m = \frac{y-y_0}{x-x_0}.
\end{equation*}
Rewriting, we obtain the important point-slope form of the line
It would not be unusual to rewrite the last equation by solving for \(y\text{.}\) In the case of the previous example, we would have
\begin{align*}
y - 2 \amp= -\frac{11}{10}(x-7)\\
y \amp= -\frac{11}{10}(x-7) + 2\\
y \amp= -\frac{11}{10}x+\frac{77}{10} + 2\\
y \amp= -\frac{11}{10}x+\frac{77}{10} + \frac{20}{10}\\
y \amp= -\frac{11}{10}x+\frac{97}{10}\\
y \amp= -1.1 x+9.7
\end{align*}
In this final form we can easily see two things: the slope \(m = -1.1\) and the location of the \(y\)-intercept, for when \(x=0\text{,}\)\(y = 9.7\text{.}\) A linear equation in the form
\begin{equation}
y = mx + b\tag{6.5.2}
\end{equation}
is said to be in slope-intercept form. The slope of the line is \(m\) and the \(y\)-intercept is \((0,b)\text{.}\)
The line has \(y\)-intercept at \((0,1)\) and slope \(m=3\text{.}\) Starting at the intercept, an increase of \(x\) by one unit will result in an increase of \(y\) by three units.
