#### Answer

(a) slope-intercept form
$\color{blue}{y=-\frac{1}{3}x+\frac{10}{3}}$
(b) standard form
$\color{red}{x+3y=10}$

#### Work Step by Step

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:
$y=-\frac{1}{3}x+\frac{10}{3}
\\y+\frac{1}{3}x=-\frac{1}{3}x+\frac{10}{3}+\frac{1}{3}x
\\\frac{1}{3}x+y=\frac{10}{3}
\\3(\frac{1}{3}x+y)=3(\frac{10}{3})
\\\color{red}{x+3y=10}$