Intermediate Algebra (12th Edition)

$y=2$
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 $( -2,2 )$ and $( 4,2 )$ is \begin{array}{l}\require{cancel} y-2=\dfrac{2-2}{-2-4}(x-(-2)) \\\\ y-2=\dfrac{2-2}{-2-4}(x+2) \\\\ y-2=\dfrac{0}{-6}(x+2) \\\\ y-2=0 \\\\ y=2 .\end{array}