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Section A.5 Finding a Point on a Line
Once we have an equation for a line (especially in point-slope form) we can easily find any point on that line by substituting in a value for
\(x\) and solving for
\(y\text{.}\) This is a basic but important skill when working with solutions to differential equations. For example, when using Euler’s method or interpreting a slope field, we often want to know the value of the solution at a particular
\(x\) -value.
🌌 Example 337 Finding a Point Given \(x\) .
The line \(L\) is given by the equation:
\begin{equation*}
y - 3 = -2(x - 5)
\end{equation*}
Find the coordinates of the point on \(L\) with \(x = 6\text{.}\)
Solution .
Substitute \(x = 6\) into the equation and solve for \(y\text{:}\)
\begin{align*}
y - 3 \amp = -2(6 - 5)\\
y - 3 \amp = -2\\
y \amp = 1
\end{align*}
So the desired point is
\((6, 1)\text{.}\)
Use the given equation of a line in point-slope form to find the point on the line with the specified
\(x\) -value.
Exercises Exercises
1.
\(y - 1 = 3(x - 9)\) \(x = 2\)
2.
\(y + 3 = \frac{1}{2}(x - 1)\text{,}\) \(x = 2\)
3.
\(y + 1 = 7x\text{,}\) \(x = -1\)
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