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Section A.4 Equation of a Line in Point-Slope Form
You’ve likely encountered the familiar slope-intercept form
\(y = mx + b\) for the equation of a line. While this form is helpful, there’s another version that plays a more direct role in differential equations: the
point-slope form . It’s especially useful when we know a point on the line and the slope, which is exactly the situation we encounter in slope fields and Euler’s method.
Recall the formula for the slope between two points, \((x_1, y_1)\) and \((x, y)\text{:}\)
\begin{equation*}
m = \frac{y - y_1}{x - x_1}
\end{equation*}
Rearranging this gives the point-slope form of a line:
\begin{equation*}
y - y_1 = m(x - x_1)
\end{equation*}
This form is powerful when modeling rates of change. For instance, when approximating solutions to differential equations numerically, each step follows the slope of a tangent line—exactly what this formula captures. You can also solve for \(y\) if needed:
\begin{equation*}
y = m(x - x_1) + y_1
\end{equation*}
🌌 Example 336 Writing a Line from Two Points.
Given the points \((3, 7)\) and \((-1, 2)\text{:}\)
Write the equation of the line in point-slope form.
Solve the equation for \(y\text{.}\)
Solution .
\begin{align*}
m \amp = \frac{2 - 7}{-1 - 3} = \frac{-5}{-4} = \frac{5}{4}
\end{align*}
Now plug into the point-slope form using either point:
\begin{align*}
y - 7 \amp = \frac{5}{4}(x - 3) \quad \text{or} \quad y - 2 = \frac{5}{4}(x + 1)
\end{align*}
\begin{align*}
y \amp = \frac{5}{4}(x - 3) + 7 \quad \text{or} \quad y = \frac{5}{4}(x + 1) + 2
\end{align*}
Exercises Practice Writing Equations of Lines
For each pair of points below:
Write the equation of the line in point-slope form.
Solve the equation for \(y\text{.}\)
1.
\((6, -1)\) and
\((2, -9)\)
2.
\((5, 0)\) and
\((1, 2)\)
3.
\((-3, -7)\) and
\((0, 4)\)
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