Chapter 3 - Introduction to Graphing - 3.7 Point-Slope Form and Equations of Lines - 3.7 Exercise Set - Page 218: 11

$\text{slope: } -\dfrac{10}{3} ,\\\text{a point on the line: } (-4,6)$

Using $y-y_1=m(x-x_1)$ or the Point-Slope form where $m$ is the slope and $(x_1,y_1)$ is a point on the line, then the given equation, $ y-6=-\dfrac{10}{3}(x+4) $, has the following characteristics: \begin{array}{l} \text{slope: } -\dfrac{10}{3} ,\\\text{a point on the line: } (-4,6) .\end{array}