## Precalculus (6th Edition)

(a) slope-intercept form $\color{blue}{y=-\frac{1}{3}x+\frac{10}{3}}$ (b) standard form $\color{red}{x+3y=10}$
RECALL: (1) The slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the line's y-intercept. (2) The standard form of a line's equation is $Ax+By=C$ where $A\ge0$ and A, B, and C are integers. (3) 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. (4) The slope $m$ of the line that contains the points $(x_1. y_1)$ and $(x_2, y_2)$ is given by the formula: $$m=\dfrac{y_2-y_1}{x_2-x_1}$$ Solve for the slope of the line that contains the two given points to obtain: $$m=\dfrac{4-3}{-2-1}=\dfrac{1}{-3}=-\dfrac{1}{3}$$ The given line has a slope of $-\frac{1}{3}$ and contains the point $(1, 3)$. This means the point-slope form of the line's equation is: $$y-3=-\frac{1}{3}(x-1)$$ (a) slope-intercept form The slope-intercept form of the line can be derived from the equation above: $y-3=-\frac{1}{3}(x-1) \\y-3=-\frac{1}{3}(x)-(-\frac{1}{3})(1) \\y-3=-\frac{1}{3}x+\frac{1}{3} \\y-3+3=-\frac{1}{3}x+\frac{1}{3}+3 \\y=-\frac{1}{3}x+\frac{1}{3}+\frac{9}{3} \\\color{blue}{y=-\frac{1}{3}x+\frac{10}{3}}$ (b) standard form The standard form of the line's equation can be derived from the slope-intercept form: $y=-\frac{1}{3}x+\frac{10}{3} \\y+\frac{1}{3}x=-\frac{1}{3}x+\frac{10}{3}+\frac{1}{3}x \\\frac{1}{3}x+y=\frac{10}{3} \\3(\frac{1}{3}x+y)=3(\frac{10}{3}) \\\color{red}{x+3y=10}$