Answer: By the slope formula.
Step-by-step explanation:
Given: ABC is a triangle (shown below),
In which A≡(6,8), B≡(2,2) and C≡(8,4)
And, D and E are the mid points of the line segments AB and BC respectively.
Prove: DE║AC and DE = AC/2
Proof:
Since, And, D and E are the mid points of the line segments AB and BC respectively.
Therefore, By mid point theorem,
coordinate of D are ![(\frac{2+6}{2} , \frac{2+8}{2} ) = (\frac{8}{2} , \frac{10}{2} )= (4,5)](https://tex.z-dn.net/?f=%28%5Cfrac%7B2%2B6%7D%7B2%7D%20%2C%20%5Cfrac%7B2%2B8%7D%7B2%7D%20%29%20%3D%20%28%5Cfrac%7B8%7D%7B2%7D%20%2C%20%5Cfrac%7B10%7D%7B2%7D%20%29%3D%20%284%2C5%29)
Coordinate of E are ![(\frac{2+8}{2} , \frac{2+4}{2} ) = (\frac{10}{2} , \frac{6}{2} )= (5,3)](https://tex.z-dn.net/?f=%28%5Cfrac%7B2%2B8%7D%7B2%7D%20%2C%20%5Cfrac%7B2%2B4%7D%7B2%7D%20%29%20%3D%20%28%5Cfrac%7B10%7D%7B2%7D%20%2C%20%5Cfrac%7B6%7D%7B2%7D%20%29%3D%20%285%2C3%29)
By the distance formula,
![DE=\sqrt{(5-4)^2+(3-5)^2}=\sqrt{5}](https://tex.z-dn.net/?f=DE%3D%5Csqrt%7B%285-4%29%5E2%2B%283-5%29%5E2%7D%3D%5Csqrt%7B5%7D)
![AC=\sqrt{(8-6)^2+(4-8)^2}=2\sqrt{5}](https://tex.z-dn.net/?f=AC%3D%5Csqrt%7B%288-6%29%5E2%2B%284-8%29%5E2%7D%3D2%5Csqrt%7B5%7D)
By the slope formula,
Slope of AC = ![\frac{4-8}{8-6} = \frac{-4}{2} = -2](https://tex.z-dn.net/?f=%5Cfrac%7B4-8%7D%7B8-6%7D%20%3D%20%5Cfrac%7B-4%7D%7B2%7D%20%3D%20-2)
Slope of DE = ![\frac{3-5}{5-4} = \frac{-2}{1} = -2](https://tex.z-dn.net/?f=%5Cfrac%7B3-5%7D%7B5-4%7D%20%3D%20%5Cfrac%7B-2%7D%7B1%7D%20%3D%20-2)
Statement Reason
1. The coordinate of D are (4,5) and 1. By the midpoint formula
the coordinate of E are (5,3)
2. The length of DE = √5 2. By the Distance formula
The length AC = 2√5 ⇒ Segment DE
is half the length of segment AC
3. The slope of DE = -2 and the 3. By the slope formula
slope of AC = -2
4. DE║AC 4. Slopes of parallel lines are equal