[tex]\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ X(\stackrel{x_1}{x}~,~\stackrel{y_1}{y})\qquad Z(\stackrel{x_2}{11}~,~\stackrel{y_2}{-5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{11+x}{2}~~,~~\cfrac{-5+y}{2} \right)=\stackrel{\stackrel{midpoint}{y}}{(3,-1)}\implies \begin{cases} \cfrac{11+x}{2}=3\\[1em] 11+x=6\\ \boxed{x=-5}\\ \cline{1-1} \cfrac{-5+y}{2}=-1\\[1em] -5+y=-2\\ \boxed{y=3} \end{cases}[/tex]
We define midpoint formula as
[tex]Y(x_m, y_m)=Y(\dfrac{x_2+x_1}{2}, \dfrac{y_2+y_1}{2})[/tex]
Also here are the coordinates of every point in variables so you won't get confused.
[tex]
Y(x_m, y_m) \\
X(x_1, y_1) \\
Z(x_2, y_2)
[/tex]
Which means there are two equations. One to find x of point X and one to find y of point X.
[tex]
x_m=\dfrac{x_2+x_1}{2}\Longrightarrow 3=\dfrac{11+x_1}{2} \\
6=11+x_1\Longrightarrow\underline{x_1=-5} \\ \\
y_m=\dfrac{y_2+y_1}{2}\Longrightarrow-1=\dfrac{-5+y_1}{2} \\
-2=-5+y_1\Longrightarrow\underline{y_1=3}
[/tex]
So point X has coordinates: [tex]\boxed{X(-5, 3)}[/tex]
Hope this helps.
r3t40
