Question

Find an equation for the perpendicular bisector of the line segment whose endpoints are (2,6) and (6,2).

Answer 1

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

`(\stackrel{x_1}{2}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{2}-\stackrel{y1}{6}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{2}}} \implies \cfrac{ -4 }{ 4 } \implies - 1 \\\\[-0.35em] ~\dotfill`

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

so the line has a slope of 1 hmmm, well, we also know is a bisector, that means it cuts the line segment into two equal halves, that means it passes through the midpoint of the line segment above, hmmm

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{2}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 6 +2}{2}~~~ ,~~~ \cfrac{ 2 +6}{2} \right) \implies \left(\cfrac{ 8 }{2}~~~ ,~~~ \cfrac{ 8 }{2} \right)\implies (4~~,~~4)`

so we're really looking for the equation of a line whose slope is 1 and it passes through (4 , 4)

`(\stackrel{x_1}{4}~,~\stackrel{y_1}{4})\hspace{10em} \stackrel{slope}{m} ~=~ 1 \\\\\\ \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}{4}=\stackrel{m}{ 1}(x-\stackrel{x_1}{4}) \\\\\\ y-4=x-4\implies {\Large \begin{array}{llll} y=x \end{array}}`