Question

If the parallel sides of a trapezoid are contained by the lines y=-1/4x+5 and y=-1/4x-1, find the equation of the line that contains the midsegment.

Answer 1

`\bf \textit{midsegment of a trapezoid}\\\\ m=\cfrac{a+b}{2}\quad \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ ----------\\ a=-(1)/(4)x+5\\\\ b=-(1)/(4)x-1 \end{cases} \\\\\\ m=\cfrac{\left( \stackrel{a}{-(1)/(4)x+5} \right)~~+~~\left(\stackrel{b}{-(1)/(4)x-1} \right)}{2} \\\\\\`


`\bf m=\cfrac{\left( (-x+20)/(4) \right)~~+~~\left((-x-4)/(4) \right)}{2}\implies m=\cfrac{(-x+20-x-4)/(4)}{2} \\\\\\ m=\cfrac{(-2x+16)/(4)}{2}\implies m=\cfrac{(-x+8)/(2)}{2}\implies m=\cfrac{(-x+8)/(2)}{(2)/(1)}\implies m=\cfrac{-x+8}{2}\cdot \cfrac{1}{2} \\\\\\ m=\cfrac{-x+8}{4}\implies m=\cfrac{8-x}{4}`