Answer
A--III
B--II
C--I
Work Step by Step
Graph I has a line that goes through the points (0,0) and (3,1). The slope of the line is $\frac{1-0}{3-0}$, or $\frac{1}{3}$. The equation of a line is y=mx+b, and we can substitute the slope and the point (0,0) into the equation to solve for b.
$ 0 = \frac{1}{3}*0 + b$
$ 0 = 0 + b$
$ 0 = b$
Thus, graph I has the equation y = 1/3 * x.
Graph II has a line that goes through the points (0,1) and (-3,0). The slope of the line is $\frac{1-0}{0--3}$, or $\frac{1}{3}$. The equation of a line is y=mx+b, and we can substitute the slope and the point (0,1) into the equation to solve for b.
$ y = mx + b $
$ 1 = 1/3 * 0 + b$
$1 = 0 + b$
$1 = b$
Thus, graph II has the equation y = 1/3 * x + 1.
By process of elimination, graph III has the equation y = 1/3 * x -1.
