The correct answer is: (-1, 3.5) and (-1, 2)
Explanation:
First let's find the equation of DF (for the sake of understanding, consider it a line).
Point D = (1,4)
Point F = (-3,3)
The slope of DF is: [tex]\frac{y_2-y_1}{x_2-x_1} = \frac{3-4}{-3-1} = \frac{-1}{-4} = \frac{1}{4}[/tex]
The equation of line is: y = mx + c
Where m = slope = [tex] \frac{1}{4}[/tex]
Equation for DF will become: [tex]y = \frac{1}{4}x + c[/tex]
To find c, plug any point (D or F) in the above equation. I am going to choose D.
4 = (1/4) + c
c = 3.75
So the equation for DF will become: y = (1/4)x + 3.75 --- (A)
Likewise, let's find the equation for EF (again assume that it is a line).
Point E = (1,1)
Point F = (-3,3)
The slope of EF is: [tex]\frac{y_2-y_1}{x_2-x_1} = \frac{3-1}{-3-1} = \frac{2}{-4} = \frac{-1}{2}[/tex]
The equation of line is: y = mx + c
Where m = slope = [tex] \frac{-1}{2}[/tex]
Equation for EF will become: [tex]y = \frac{-1}{2}x + c[/tex]
To find c, plug any point (E or F) in the above equation. I am going to choose E.
1 = (-1/2) + c
c = 1.5
So the equation for EF will become: y = (-1/2)x + 1.5 --- (B)
We can see in the graph that the midsegment that is parallel to DE will be at x = -1 (two unit to the left of DE), since the △DEF is 4 units wide.
Put x = -1 in equation (A) to get the upper endpoint of the midsegment:
y = (1/4)(-1) + 3.75
y = 3.5 (when x = -1) <-- Upper endpoint
Put x = -1 in equation (B) to get the lower endpoint of the midsegment:
y = (-1/2)(-1) + 1.5
y = 2 (when x = -1) <-- lower endpoint
Hence, (-1, 3.5) and (-1, 2) is the correct answer.
