Question

Find an equation for the perpendicular bisector of the line segment whose endpoints are (7,5)and(−1,9).

Answer 1

Check the picture below.

so we're really looking for the equation of the red line hmmm, so hmmm is a bisector, that means it passes through the midpoint of that segment, so

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{7}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{9}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +7}{2}~~~ ,~~~ \cfrac{ 9 +5}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 14 }{2} \right)\implies (3~~,~~7)`

now, hmmm keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the segment above

`(\stackrel{x_1}{7}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{9}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{9}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{-1}-\underset{x_1}{7}}} \implies \cfrac{4}{-8}\implies -\cfrac{1}{2}`

now let's find the slope of a perpendicular to that one

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-1}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{-1}\implies 2}}`

so, we're really looking for the equation of a line that passes through the midpoint (3 , 7) and that it has a slope of 2

`(\stackrel{x_1}{3}~,~\stackrel{y_1}{7})\hspace{10em} \stackrel{slope}{m} ~=~ 2 \\\\\\ \begin{array}c \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}{7}=\stackrel{m}{ 2}(x-\stackrel{x_1}{3}) \\\\\\ y-7=2x-6\implies {\LARGE \begin{array}{llll} y=2x+1 \end{array}}`