## Intermediate Algebra (12th Edition)

$y=-6$
Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the two-point form of linear equations, the equation of the line passing through $\left( -\dfrac{4}{9},-6 \right)$ and $\left( \dfrac{12}{7},-6 \right)$ is \begin{array}{l}\require{cancel} y-(-6)=\dfrac{-6-(-6)}{-\frac{4}{9}-\frac{12}{7}}\left( x-\left(-\dfrac{4}{9}\right) \right) \\\\ y+6=\dfrac{-6+6}{-\frac{4}{9}-\frac{12}{7}}\left( x+\dfrac{4}{9}\right) \\\\ y+6=\dfrac{0}{-\frac{4}{9}-\frac{12}{7}}\left( x+\dfrac{4}{9}\right) \\\\ y+6=0 \\\\ y=-6 .\end{array}