Answer:
1 and 2.
Midpoints calculated, plotted and connected to make the triangle DEF, see the attached.
- D= (-2, 2), E = (-1, -2), F = (-4, -1)
3.
As per definition, midsegment is parallel to a side.
Parallel lines have same slope.
Find slopes of FD and CB and compare.
- m(FD) = (2 - (-1))/(-2 -(-4)) = 3/2
- m(CB) = (1 - (-5))/(1 - (-3)) = 6/4 = 3/2
- As we see the slopes are same
Find the slopes of FE and AB and compare.
- m(FE) = (-2 - (- 1))/(-1 - (-4)) = -1/3
- m(AB) = (1 - 3)/(1 - (-5)) = -2/6 = -1/3
- Slopes are same
Find the slopes of DE and AC and compare.
- m(DE) = (-2 - 2)/(-1 - (-2)) = -4/1 = -4
- m(AC) = (-5 - 3)/(-3 - (-5)) = -8/2 = -4
- Slopes are same
4.
As per definition, midsegment is half the parallel side.
We'll show that FD = 1/2CB
- FD = [tex]\sqrt{(2+1)^2+(-2+4)^2}[/tex] = [tex]\sqrt{3^2+2^2}[/tex] = [tex]\sqrt{13}[/tex]
- CB = [tex]\sqrt{(1 + 5)^2+(1+3)^2}[/tex] = [tex]\sqrt{6^2+4^2}[/tex] = 2[tex]\sqrt{13}[/tex]
- As we see FD = 1/2CB
FE = 1/2AB
- FE = [tex]\sqrt{(-4+1)^2+(-1+2)^2}[/tex] = [tex]\sqrt{3^2+1^2}[/tex] = [tex]\sqrt{10}[/tex]
- AB = [tex]\sqrt{(-5 -1)^2+(3-1)^2}[/tex] = [tex]\sqrt{6^2+2^2}[/tex] = 2[tex]\sqrt{10}[/tex]
- As we see FE = 1/2AB
DE = 1/2AC
- DE = [tex]\sqrt{(-2+1)^2+(2+2)^2}[/tex] = [tex]\sqrt{1^2+4^2}[/tex] = [tex]\sqrt{17}[/tex]
- AC = [tex]\sqrt{(-5 +3)^2+(3+5)^2}[/tex] = [tex]\sqrt{2^2+8^2}[/tex] = 2[tex]\sqrt{17}[/tex]
- As we see DE = 1/2AC
