A triangle has vertices at (4, 4), (-6, 2) and (2, 0).

a. Find the coordinates of the mid-points of eachside

b. Find the lengths of the sides of the triangle
formed by joining the mid-points.
each side.

Answer:

A. (-1, 3) (3, 2) (-2, 1)

B. [tex]\sqrt{17}[/tex]   [tex]\sqrt{5}[/tex]  [tex]\sqrt{26}[/tex]  

Step-by-step explanation:

The formula for finding the midpoints is [tex](\frac{x_{1}+x_{2} }2}, \frac{y_{1}+y_{2} }2} )[/tex].

MIDPOINTS

[tex](\frac{x_{1}+x_{2} }2}, \frac{y_{1}+y_{2} }2} )[/tex] = [tex](\frac{4-6}2,\frac{4+2 }2} )[/tex] = (-1, 3)

[tex](\frac{x_{1}+x_{2} }2}, \frac{y_{1}+y_{2} }2} )[/tex] = [tex](\frac{4+2}2,\frac{4+0 }2} )[/tex] = (3, 2)

[tex](\frac{x_{1}+x_{2} }2}, \frac{y_{1}+y_{2} }2} )[/tex] = [tex](\frac{-6+2}2,\frac{2+0 }2} )[/tex] = (-2, 1)

Next, we will use the distance formula, [tex]\sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2} }[/tex].

DISTANCE

[tex]\sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2} }[/tex] = [tex]\sqrt{(-1-3)^{2} + (3-2)^{2} }[/tex] = [tex]\sqrt{17}[/tex]

[tex]\sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2} }[/tex] = [tex]\sqrt{(-1+2)^{2} + (3-1)^{2} }[/tex] = [tex]\sqrt{5}[/tex]

[tex]\sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2} }[/tex] = [tex]\sqrt{(3+2)^{2} + (2-1)^{2} }[/tex] = [tex]\sqrt{26}[/tex]  

a. To find the coordinates of endpoints we must add two x values and divide by 2 and then add 2 y- values and divide by 2.

(4-6)/2=-1  (4+2)/2=3

Repeat for other sides.

(2-6)/2=-2 (2+0)/2=1

(2+4)/2=3  (4+0)/2=2

Coordinates of midpoints are (-1,3), (-2,1), (3,2)

b. Now we use the distance formula for each midpoint to find the length of the inner- triangle.

sqrt((-1+2)^2 + (3-1)^2)

Sqrt(5)

Repeat.

Sqrt(17)

Sqrt(26)

The lengths of the inner triangle are as follows:

(-1,3), (-2,1) = Sqrt(5)

(-1,3), (3,2) = Sqrt(17)

(-2,1), (3,2) = Sqrt(26)

anlatarak cozmicekseniz cevaplamayin:) ​

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