Question

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

Answer 1

Answer:

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

B. $\sqrt{17}$   $\sqrt{5}$  $\sqrt{26}$  

Step-by-step explanation:

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

MIDPOINTS

$(\frac{x_{1}+x_{2} }2}, \frac{y_{1}+y_{2} }2} )$ = $(\frac{4-6}2,\frac{4+2 }2} )$ = (-1, 3)

$(\frac{x_{1}+x_{2} }2}, \frac{y_{1}+y_{2} }2} )$ = $(\frac{4+2}2,\frac{4+0 }2} )$ = (3, 2)

$(\frac{x_{1}+x_{2} }2}, \frac{y_{1}+y_{2} }2} )$ = $(\frac{-6+2}2,\frac{2+0 }2} )$ = (-2, 1)

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

DISTANCE

$\sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2} }$ = $\sqrt{(-1-3)^{2} + (3-2)^{2} }$ = $\sqrt{17}$

$\sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2} }$ = $\sqrt{(-1+2)^{2} + (3-1)^{2} }$ = $\sqrt{5}$

$\sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2} }$ = $\sqrt{(3+2)^{2} + (2-1)^{2} }$ = $\sqrt{26}$  

Answer 2

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)