Answer
$y=-\frac{1}{3}x + \frac{10}{3}$.
Work Step by Step
RECALL:
(i)
The slope-intercept form of a line's equation is $y=mx+b$ where $m$ = slope and $b$ = y-intercept.
(ii) The formula for slope is $m=\dfrac{y_2-y_1}{x_2-x_1}$.
Solve for the slope using the formula above to have:
$m=\dfrac{3-2}{1-4}=\dfrac{1}{-3}=-\dfrac{1}{3}$
Thus, the tentative equation of the line is $y=-\frac{1}{3}x + b$.
Solve for the value of $b$ by substituting the x and y-coordinates of one point into the tentative equation to have:
$y=-\frac{1}{3}x+b
\\3 = -\frac{1}{3}(1)+b
\\3=-\frac{1}{3}+b
\\3+\frac{1}{3}=b
\\\frac{10}{3}=b$
Therefore, the equation of the line in slope-intercept form is $y=-\frac{1}{3}x + \frac{10}{3}$.
