Question

Consider the line y = 5/7x+1.

Find the equation of the line that is parallel to this line and passes through the point (-5, 6).
Find the equation of the line that is perpendicular to this line and passes through the point (-5, 6).
Note that the ALEKS graphing calculator may be helpful in checking your answer.
Equation of parallel line:
Equation of perpendicular line:
П
X√√
0=0
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Answer 1

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

`y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{5}{7}}x+1\qquad \impliedby \begin{array} \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}`

so for the parallel line we're really lookikng for the equation of aline whose slope is 5/7 and that it passes through (-5 , 6)

`(\stackrel{x_1}{-5}~,~\stackrel{y_1}{6})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{5}{7} \\\\\\ \begin{array} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{6}=\stackrel{m}{ \cfrac{5}{7}}(x-\stackrel{x_1}{(-5)}) \implies y -6= \cfrac{5}{7} (x +5) \\\\\\ y-6=\cfrac{5}{7}x+\cfrac{25}{7}\implies y=\cfrac{5}{7}x+\cfrac{25}{7}+6\implies {\Large \begin{array}{llll} y=\cfrac{5}{7}x+\cfrac{67}{7} \end{array}}`

now, keeping in mind that perpendicular lines have negative reciprocal slopes

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{5}{7}} ~\hfill \stackrel{reciprocal}{\cfrac{7}{5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{7}{5}}}`

so for the perpendicular line we're really looking for the equation of a line whose slope is -7/5 and that it passes through (-5 , 6)

`(\stackrel{x_1}{-5}~,~\stackrel{y_1}{6})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{7}{5} \\\\\\ \begin{array} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{6}=\stackrel{m}{- \cfrac{7}{5}}(x-\stackrel{x_1}{(-5)}) \implies y -6= -\cfrac{7}{5} (x +5) \\\\\\ y-6=-\cfrac{7}{5}x-7\implies {\Large \begin{array}{llll} y=-\cfrac{7}{5}x-1 \end{array}}`