# Chapter 2 - Section 2.3 - Lines - 2.3 Assess Your Understanding - Page 178: 42

$\color{blue}{y=\dfrac{1}{3}x+\dfrac{4}{3}}$

#### Work Step by Step

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 can be solved using the formula $m=\dfrac{y_2-y_1}{x_2-x_1}$, where $(x_1, y_1)$ and $(x_2, y_2)$ are points on the line. The given line contains the points $(-1, 1)$ and $(2, 2)$. Solve for the slope using the formula in $(2)$ above to obtain: $m=\dfrac{1-2}{-1-2} \\m=\dfrac{-1}{-3} \\m=\dfrac{1}{3}$ Thus, the tentative equation of the line is: $y=\dfrac{1}{3}x+b$ To find the value of $b$, substitute the x and y values of the point $(2, 2)$ into the tentative equation above to obtain: $y=\dfrac{1}{3}x+b \\2=\dfrac{1}{3}(2)+b \\2=\dfrac{2}{3}+b \\2-\dfrac{2}{3} = b \\\dfrac{6}{3} - \dfrac{2}{3}=b \\\dfrac{4}{3} = b$ Therefore, the equation of the line is: $\color{blue}{y=\dfrac{1}{3}x+\dfrac{4}{3}}$

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