Question

For a (1,3) and b (1,5), what is the equation of the perpendicular bisector of ab

Answer 1

bearing in mind that perpendicular lines have negative reciprocal slopes, hmmmm what is the slope of AB anyway?


`\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad B(\stackrel{x_2}{1}~,~\stackrel{y_2}{5}) \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-3}{1-1}\implies \stackrel{und efined}`

so, AB has an undefined slope of 2/0,


`\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{2}{0}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{0}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{0}{2}}\implies 0}`

so it has a slope of 0 then, and we also know is a bisector of AB, meaning it passes AB through its midpoint, since it cuts it in two equal halves.


`\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad B(\stackrel{x_2}{1}~,~\stackrel{y_2}{5}) \qquad % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{1+1}{2}~~,~~\cfrac{5+3}{2} \right)\implies (1,4)`

so, we're really looking for the equation of a line whose slope is 0, and runs through 1,4.


`\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad \qquad \qquad slope = m\implies 0 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-4=0(x-1)\implies y-4=0\implies y=4`