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.

keeping in mind that perpendicular lines have negative reciprocal slopes, and that we also know that the line is a perpendicular bisector, namely it cuts that segment into two equal halves as you can see in the picture, so it really passes through the midpoint of that segment, so let's get the slope of that green segment

`(\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}`

so for the perpendicular bisector that'd be

`\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}}`

now let's check for the midpoint of the green segment

`~~~~~~~~~~~~\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)`

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

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