Question

L = (15,15) M = (13,5) N= (-3,-3) What is the equation of the median from vertex N to LM?

Answer 1

we know the median stems out of vertex N and on the half-way between L and M, so let's find the midpoint of LM firstly.

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{13}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 13 + 15}{2}~~~ ,~~~ \cfrac{ 5 + 15}{2} \right)\implies \left(\cfrac{28}{2}~~,~~\cfrac{20}{2} \right)\implies (14~~,~~10)$

so we're really looking for the equation of a line that goes through (14,10) and (-3,-3), Check the picture below.

$(\stackrel{x_1}{14}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{-3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{-3}-\underset{x_1}{14}}}\implies \cfrac{-13}{-17}\implies \cfrac{13}{17}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{\cfrac{13}{17}}(x-\stackrel{x_1}{14}) \\\\\\ y-10=\cfrac{13}{17}x-\cfrac{182}{17}\implies y=\cfrac{13}{17}x-\cfrac{182}{17}+10\implies y=\cfrac{13}{17}x-\cfrac{12}{17}$