Question

Find the equation of the line that is parallel to the line 0%2B%204" id="TexFormula1" title="y = -\frac{3}{2}x + 4" alt="y = -\frac{3}{2}x + 4" align="absmiddle" class="latex-formula"> and passes through the point (4, 0)
Work needs to be shown

Answer 1

parallel lines have the same exact slope, so

$\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{7em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{3}{4}}x+4}$

so we're really looking for a line whose slope is -3/4 and runs through (4,0),

$\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{0})~\hspace{10em} slope = m\implies -\cfrac{3}{4} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-0=-\cfrac{3}{4}(x-4)\implies y=-\cfrac{3}{4}x+3$

Answer 2

A line parallel to your equation has the same slope, so it should be in the form:

y = (-3/2)x + b

To figure what "b" has to be, plug in the point (4,0) and solve:

0 = (-3/2)*4 + b

0 = -6 + b

6 = b

So the equation of the line is:

y = (-3/2)x + 6