Find the midpoint and distance between (5,1) and (-15,11)
1 answer:
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Answer:
AC = 6
Step-by-step explanation:
y is the dimension of the horizontal segment (see the attached image).
The hypotenuses are the same dimension, so:
(x+4)^2=(x/2)^2+y^2
(3x-8)^2=(x/2)^2+y^2
So,
(x+4)^2 = (3x-8)^2
x+4 = 3x-8
x-3x = -8-4
-2x = -12
x = 6
And x is the dimension of the segment AC.
