Respuesta :
The midpoint of a line segment is just the average of the coordinates of the endpoints.
(0+x)/2=-3 and (2+y)/2=2
x=-6 and 2+y=4
x=-6 and y=2
So point B is (-6,2)
(0+x)/2=-3 and (2+y)/2=2
x=-6 and 2+y=4
x=-6 and y=2
So point B is (-6,2)
Answer:
A= (-6, 2)
Step-by-step explanation:
Let ([tex]x_{1}[/tex] [tex]y_{1}[/tex]) be the coordinates of A , let ( [tex]x_{2}[/tex] , [tex]y_{2}[/tex]) be the coordinate of B
and let ( [tex]x_{m}[/tex] , [tex]y_{m}[/tex]) be the midpoints of segment AB
To find the coordinates of A, we simply use the formula for calculating mid-points of segment
( [tex]x_{m}[/tex] , [tex]y_{m}[/tex]) = [tex](\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2})[/tex]
From the formula above;
[tex]x_{m}[/tex] = [tex]x_{1}[/tex] + [tex]x_{2}[/tex] / 2 ---------------(1)
Similarly
[tex]y_{m}[/tex] = [tex]y_{1}[/tex] + [tex]y_{2}[/tex] / 2 --------(2)
From the question given;
( [tex]x_{m}[/tex] , [tex]y_{m}[/tex]) = (-3, 2) which implies: [tex]x_{m}[/tex]= -3 and [tex]y_{m}[/tex]= 2
similarly ( [tex]x_{2}[/tex] , [tex]y_{2}[/tex]) = (0,2) this implies that [tex]x_{2}[/tex] = 0 and [tex]y_{2}[/tex] =2
From equation (1)
[tex]x_{m}[/tex] = [tex]x_{1}[/tex] + [tex]x_{2}[/tex] / 2
we substitute [tex]x_{m}[/tex]= -3 and [tex]x_{2}[/tex] = 0 into equation (1) to get [tex]x_{1}[/tex]
[tex]x_{m}[/tex] = [tex]x_{1}[/tex] + [tex]x_{2}[/tex] / 2
-3 = [tex]x_{1}[/tex] + 0 / 2
cross multiply
[tex]x_{1}[/tex] + 0 = -3 × 2
[tex]x_{1}[/tex] = -6
Also;
We substitute; [tex]y_{m}[/tex]= 2 and [tex]y_{2}[/tex] =2 into equation (2)
[tex]y_{m}[/tex] = [tex]y_{1}[/tex] + [tex]y_{2}[/tex] / 2 --------(2)
2 = [tex]y_{1}[/tex] + 2 / 2
cross multiply
[tex]y_{1}[/tex] + 2 = 2 × 2
[tex]y_{1}[/tex] + 2 = 4
subtract 2 from both-side of the equation
[tex]y_{1}[/tex] + 2 -2 = 4 -2
[tex]y_{1}[/tex] = 2
[tex]x_{1}[/tex] = -6 and [tex]y_{1}[/tex] = 2
Therefore; Coordinate of A ([tex]x_{1}[/tex] [tex]y_{1}[/tex]) = (-6, 2)
