Answer
point-slope form: $y=\frac{1}{2}(x)$
slope-intercept form: $y= \frac{1}{2}x$
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 point-slope form of a line's equation is:
$y-y_1=m(x-x_1)$
(a) point-slope form
The given line has a slope of $\frac{1}{2}$ and passes through the origin, $(0, 0)$.
Substitute these values into the point-slope form above to obtain:
$y-0=\frac{1}{2}(x-0)
\\y = \frac{1}{2}(x)$
(b) slope-intercept form
Substitute the slope $\frac{1}{2}$ to $m$ to obtain the tentative equation:
$y=\frac{1}{2}x+b$
The line passes through the origin, $(0, 0)$.
This means that the coordinates of this point satisfy the equation of the line. Substitute the x and y-coordinates of this point into the tentative equation to obtain:
$y=\frac{1}{2}x+b
\\0 = \frac{1}{2}(0) + b
\\0 = 0 + b
\\0= b$
Thus, the equation of the line is $y=\frac{1}{2}x$.
