## College Algebra (10th Edition)

$\color{blue}{y=\dfrac{1}{2}x+\dfrac{5}{2}}$
RECALL: (1) The slope-intercept form of a line's equation is: $y=mx+b$, where $m$=slope and $b$ = y-intercept. (2) The slope of a line that contains the points $(x_1)$ and $(x_2, y_2)$ can be found using the formula: $m=\dfrac{y_2-y_1}{x_2-x_1}$ Solve for the slope of the line using the two given points on the line to obtain: $m=\dfrac{2-3}{-1-1} \\m=\dfrac{-1}{-2} \\m=\dfrac{1}{2}$ Thus, the tentative equation of the line is: $y=\dfrac{1}{2}x+b$ To find the value of $b$, substitute the x and y values of the point $(1, 3)$ into the tentative equation above to obtain: $y=\dfrac{1}{2}x+b \\3=\dfrac{1}{2}(1)+b \\3=\dfrac{1}{2}+b \\3-\dfrac{1}{2}=b \\\dfrac{6}{2}-\dfrac{1}{2}=b \\\dfrac{5}{2}=b$ Therefore, the equation of the line is: $\color{blue}{y=\dfrac{1}{2}x+\dfrac{5}{2}}$.