Answer :
[tex] \frac{-2+x}{2}=1 [/tex]∧[tex] \frac{3+y}{2} =0[/tex]⇔[tex]x = 4[/tex] ∧[tex]y = -3[/tex]
R: D (4, -3)
R: D (4, -3)
Answer:
The other endpoint
[tex](x_{2} , y_{2}) is (4,-3)[/tex]
Step-by-step explanation:
Given
Point 1 (-2,3)
Midpoint = (1,0)
We're to calculate the other end point.
To solve this, we'll make use of the formula of midpoints of line
Given that two lines of coordinates [tex]P(x_{1} ,y_{1} )[/tex] and [tex]Q(x_{2} ,y_{2} )[/tex]
[tex](m_{1},m_{2}) = (\frac{(x_{1} + x_{2} )}{2} ,\frac{(y_{1} + y_{2} )}{2})\\Where\\m_{1} = \frac{(x_{1} + x_{2} )}{2} \\and\\m_{2} = \frac{(y_{1} + y_{2} )}{2})\\[/tex]
From the question, we have
[tex]m_{1} = 1\\m_{2} = 0\\x_{1} = -2\\y_{1} = 3\\[/tex]
Calculating [tex]x_{2}[/tex]...
From [tex]m_{1} = \frac{(x_{1} + x_{2} )}{2}[/tex]
By Substitution, we have
[tex]1 = \frac{-2+x_{2} }{2} \\2 = -2+x_{2}\\2 + 2 = x_{2}\\ x_{2} = 4[/tex]
Calculating [tex]y_{2}[/tex]...
From [tex]m_{2} = \frac{(y_{1} + y_{2} )}{2}[/tex]
By Substitution, we have
[tex]0 = \frac{3+y_{2} }{2} \\0 = 3+y_{2}\\0 - 3 = y_{2}\\ y_{2} = -3[/tex]
Hence, the other endpoint
[tex](x_{2} , y_{2}) is (4,-3)[/tex]
