Question 214906
a. the point in the axis that is equidistant from 12,6 and 8,2
What does "in the axis" mean?
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b. the distance between the points 6,4 and 18,4
s^2 = diffy^2 + diffx^2
s^2 = 0 + (18-6)^2
s^2 = 12^2
s = 12
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c. the distance between the points 1,4 and 19,6
Do it like b above.
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d. the midpoint between 12,6 and 18,4
Find the average of x and y separately
x: (12+18)/2 = 15
y: (6+4)/2 = 5
Midpoint = (15,5)
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e. the midpoint between -6, 8 and 23, 5
Same as d above.