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Section Finding a Point on a Line
Once we have an equation for a lineβespecially in point-slope formβwe 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 Eulerβs method or interpreting a slope field, we often want to know the value of the solution at a particular
\(x\) -value.
π Example 308 . 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 \(x = 6\text{.}\)
Answer .
\begin{equation*}
(6, 1)
\end{equation*}
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. \(y - 1 = 3(x - 9)\text{,}\) ββ\(x = 2\) 2. \(y + 3 = \frac{1}{2}(x - 1)\text{,}\) ββ\(x = 2\) 3. \(y + 1 = 7x\text{,}\) ββ\(x = -1\) You have attempted
of
activities on this page.
