Answer:
1) AB=7 CD=3 EF=[tex]3\sqrt{5}[/tex] 2) [tex]M_{AB}=\frac{-1}{2},2[/tex] [tex] \\ M_{EF} =(\frac{5}{2}-3)[/tex] \\ [tex] \\ M_{EF} =(\frac{5}{2}-3)[/tex] 3) AB not inclined CD not inclined EF 6/5
Step-by-step explanation:
1) We can use the Distance Formula to answer the 1st. question.
But in the first case AB I'd rather doing it intuitively because it is a straight line parallel to the x-axis
AB
A(-4,2) and B(3,2)
In this segment, also parallel we can calculate the length as |-4|+|3|=7 since both have the same y coordinate.
Using the Distance formula to check it:
[tex]D=\sqrt{(3--4)^{2}+(2-2)^{2}}\\D=\sqrt{49}\\D=7[/tex]
CD
C(-4,-1) D(-4,-4)
Similar to the first one but this time with different y coordinates.
The length will be calculated by subtracting the absolute values for y:
|-4|-|-1|=4-1 = 3
Using the Distance formula to check it:
[tex]D=\sqrt{(-4--1)^{2}+(-4--4)^{2}}\\D=\sqrt{9}\\D=3[/tex]
EF
E (1,-1) F(4,-5)
In this case there's no straight line.
So right to the Distance Formula:
[tex]D=\sqrt{(-5-1)^{2}+(4-1)^{2}}\\ D=\sqrt{36+9}\\D=\sqrt{45}\\D=3\sqrt{5}[/tex]
2) To find the Midpoints we need to calculate the Mean of these two points.
AB
A(-4,2) and B(3,2)
[tex]M_{AB} =\frac{-4+3}{2},\frac{2+2}{2}\\M_{AB}=\frac{-1}{2},2[/tex]
CD
C(-4,-1) D(-4,-4)
[tex]M_{CD} =\frac{-4-4}{2} ,\frac{-1-4}{2} \\M_{CD} =(0,\frac{-5}{2})[/tex]
EF
E (1,-1) F(4,-5)
[tex]M_{EF} =\frac{4+1}{2} ,\frac{-1-5}{2} \\ M_{EF} =(\frac{5}{2}-3)[/tex]
3) To find the Slope let's calculate the quotient of a difference between y-coordinates over x-coordinates of two given points.
[tex]m_{AB}=\frac{2-2}{3--4}=\frac{0}{7}=0[/tex]
AB is not inclined.
CD
C(-4,-1) D(-4,-4)
[tex]m_{CD}=\frac{-4--1}{-4--4}=\frac{-3}{-4+4}=\frac{-3}{0}[/tex]
Not Defined for all Real Set of Numbers
The line CD is not inclined.
EF
E(1,-1) F(4,-5)
[tex]m_{EF}=\frac{-5-1}{4--1}=\frac{6}{5}[/tex]
The line EF has a slope of 6/5
