Question

Can someone help me pleaseeeee

Answer 1

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the line segment above, and to get that, we only need two points off of it, let's use those two in the picture below.

`(\stackrel{x_1}{-2}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{-4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-4}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{2}-\underset{x_1}{(-2)}}} \implies \cfrac{ -8 }{2 +2} \implies \cfrac{ -8 }{ 4 } \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} \implies \cfrac{1}{ 2 }}}`

so we are really looking for the equation of a line whose slope is 1/2 and it's the bisector of that segment, well, in order to bisect it, it must pass through the midpoint of that line, now, we could run a midpoint calculation, but it suffices to say is pretty much at the origin, rather very clear from the picture, so it must also pass through (0 , 0)

`(\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{2} \\\\\\ \begin{array}c \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}{0}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{0}) \\\\\\ y=\cfrac{x}{2}\implies 2y=x\implies {\Large \begin{array}{llll} 2y-x=0 \end{array}}`