## College Algebra (10th Edition)

point-slope form: $y-3=-\dfrac{2}{5}(x-1)$ Refer to the image below for the graph.
(A) To graph the line with the given point and slope (slope is $\dfrac{\text{rise}}{\text{run}}$), perform the following steps: (1) Plot the point P(1, 3). (2) Use the slope $-\dfrac{2}{5}$ to get another point on the line. From $(1, 3)$, move 2 units downward (the rise) and 5 units to the right (the run) to arrive at the point $(6, 1)$. (3) Connect the two points using a straight line to complete the graph. (refer to the attached image below) (B) The point-slope form of a line's equation is $y-y_1 = m(x-x_1)$ where $m$= slope and $(x_1, y_1)$ is a point on the line. With $m=-\dfrac{2}{5}$ and $P(1, 3)$ on the line, the point-slope form of the equation of the line is: $y-3=-\dfrac{2}{5}(x-1)$