Question

100 points! Mhanifa can you please help? Look at the picture attached. I will mark brainliest!

Answer 1

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 = $\sqrt{(2+1)^2+(-2+4)^2}$ = $\sqrt{3^2+2^2}$ = $\sqrt{13}$
• CB = $\sqrt{(1 + 5)^2+(1+3)^2}$ = $\sqrt{6^2+4^2}$ = 2$\sqrt{13}$
• As we see FD = 1/2CB

FE = 1/2AB

• FE = $\sqrt{(-4+1)^2+(-1+2)^2}$ = $\sqrt{3^2+1^2}$ = $\sqrt{10}$
• AB = $\sqrt{(-5 -1)^2+(3-1)^2}$ = $\sqrt{6^2+2^2}$ = 2$\sqrt{10}$
• As we see FE = 1/2AB

DE = 1/2AC

• DE = $\sqrt{(-2+1)^2+(2+2)^2}$ = $\sqrt{1^2+4^2}$ = $\sqrt{17}$
• AC = $\sqrt{(-5 +3)^2+(3+5)^2}$ = $\sqrt{2^2+8^2}$ = 2$\sqrt{17}$
• As we see DE = 1/2AC

Answer 2

Answer:

thanks for the points bro jk the answers are 1234

Step-by-step explanation: