Beginning Algebra With Applications, Chapter 5, 5.4, Section 5.4, Problem 62

Is the linear equation that contains all the ordered pairs $(-2,-4),(0,-3),(4,-1)$? If there is, find the equation.

To check if the given pairs has a linear equation, we find first the slope. So we let $(x_1, y_1) = (-2,-4)$ and $(x_2, y_2) = (0,-3)$,

$\displaystyle m = \frac{-3-(-4)}{0-(-2)} = \frac{-3+4}{2} = \frac{1}{2}$

Then we let $(x_1, y_1) = (0, -3)$ and $(x_2, y_2) = (4,-1)$

$\displaystyle m = \frac{-1-(-3)}{4-0} = \frac{-1+3}{4} = \frac{2}{4} = \frac{1}{2}$

And we let $(x_1, y_1) = (-2, -4)$ and $(x_2, y_2) = (4,-1)$

$\displaystyle m = \frac{-1-(-4)}{4-(-2)} = \frac{-1+4}{4+2} = \frac{3}{6} = \frac{1}{2}$

Since they have the same slope. The given ordered pairs has a linear equation

Using the point-slope formula, where $\displaystyle m = \frac{1}{2}$ and $(x_1, y_1) = (-2,-4)$


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x -x_1)
\\
\\
y - (-4) =& \frac{1}{2} [x- (-2)]
\\
\\
y + 4 =& \frac{1}{2}x + 1
\\
\\
y =& \frac{1}{2}x - 3

\end{aligned}
\end{equation}
$


The equation is $\displaystyle y = \frac{1}{2}x - 3$.