Question

Geometry help needed!

Answer 1

Answer:

AC Midpoint (2,0)-

CB Midpoint (5,-4)

midlength is 5

Step-by-step explanation:

midpoint formula $(\frac{x_{1}+x_{2} }{2} ,\frac{y_{1} +y_{2} }{2})$

A(2,4) C(2,-4)

$\frac{2+2}{2} =\frac{4}{2}=2\\\frac{4-4}{2} =\frac{0}{2} =0$

(2,0)

C(2,-4) B(8,-4)

$\frac{8+2}{2} =\frac{10}{2} =5$

$\frac{-4-4}{2} =\frac{-8}{2} =-4$  midpoint is  (5,-4)

Distance between the 2 midpoints is the midsegment

(2,0) and (5,-4)

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

$d=\sqrt{(5-2)^{2} +(-4-0)^{2} } \\d=\sqrt{3^{2} +(-4^{2})\\ } \\d=\sqrt{x} 9+16\\d=\sqrt{25} \\d=5$