Question

PLEASE HELP ME SOLVE THIS ONE QUESTION , I HAVE SOLVED I) IT IS II) I NEED HELP WITH

5. A is the point (1,5) and B is the point (3,9).M is the midpoint of AB
i) M = (2,5)
ii)Find the equation of the line that is perpendicular to AB and passes through M.
Give your answer in the form : y=mx+c

Answer 1

I)

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{3}~,~\stackrel{y_2}{9}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 +1}{2}~~~ ,~~~ \cfrac{ 9 +5}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ 14 }{2} \right)\implies (2~~,~~7)`

II)

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

`(\stackrel{x_1}{1}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{9}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{9}-\stackrel{y1}{5}}}{\underset{\textit{\large run}} {\underset{x_2}{3}-\underset{x_1}{1}}} \implies \cfrac{ 4 }{ 2 } \implies 2 \\\\[-0.35em] ~\dotfill`

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

so we're really looking for the equation of a line whose slope is -1/2 and it passes through (2 , 7)

`(\stackrel{x_1}{2}~,~\stackrel{y_1}{7})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{1}{2} \\\\\\ \begin{array}ll \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}{- \cfrac{1}{2}}(x-\stackrel{x_1}{2}) \\\\\\ y-7=- \cfrac{1}{2}x+1\implies {\Large \begin{array}{llll} y=- \cfrac{1}{2}x+8 \end{array}}`