can u pls help because it is difficult
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Answer:
C, D
Step-by-step explanation:
The equation for the x-coordinate will have the x-coordinate of the midpoint (-1/2) on the left side of an equation involving x2. Choice C applies.
The equation for the y-coordinate will have the y-coordinate of the midpoint (-1) on the left side of an equation involving y2. Choice D applies.
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Additional comment
When you solve the equation
M = (P +Q)/2
for P, you find that ...
P = 2M -Q . . . . . . . . our favorite way to find the other end point
or ...
x2 = 2(-1/2) -3 = -4
y2 = 2(-1) -(-5) = 3
The coordinates of P are (-4, 3).
[tex]~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ Q(\stackrel{x_1}{3}~,~\stackrel{y_1}{-5})\qquad P(\stackrel{x_2}{x_2}~,~\stackrel{y_2}{y_2}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ x_2 + 3}{2}~~~ ,~~~ \cfrac{ y_2 -5}{2} \right)~~ = ~~\stackrel{\textit{\Large Midpoint}}{\left(-\cfrac{1}{2}~~,~~-1 \right)} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\cfrac{ x_2 + 3}{2}~~ = ~~-\cfrac{1}{2}\implies \boxed{-\cfrac{1}{2}=\cfrac{ 3+x_2 }{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{ y_2 -5}{2}~~ = ~~-1\implies \boxed{-1=\cfrac{-5+y_2}{2}}[/tex]