Question

Find the distance between (3, 8) and (7, -11). Round to the nearest hundredth.

Answer 1

Distance formula : d = sqrt (x2 - x1)^2 + (y2 - y1)^2
(3,8)(-7,11)
d = sqrt (-7 - 3)^2 + (11 - 8)^2
d = sqrt (-10^2) + (3^2)
d = sqrt (100 + 9)
d = sqrt 109
d = 10.44
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mid-point formula : ((x1 + x2) / 2 , (y1 + y2) / 2)
(-9,5)(21,-7)

midpoint = ((-9 + 21)/2 , (5 + (-7)/2)
               = ((12/2) , (-2,2))
               = (6, -1) <==
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(-8,7)(-9,11)
midpoint = ((-8 + (-9) / 2, (7 + 11)/2 )
               = (-17/2, 18/2)
               = (-8.5, 9) <==
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(8,y)(-11,6).....midpoint = (-1.5/5)
midpoint = ((8 + (-11)/2, (y + 6)/2)
               = (-3/2), (4 + 6)/2
               = (-1.5,5)
               y = 4 <==
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slope = (y2 - y1) / (x2 - x1)
(3,-6)(6,12)
slope = (12 - (-6) / (6 - 3) = (12 + 6) / 3 = 18/3 = 6 <==
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(x,10)(-4,8)...slope = 2/3

y = mx + b
slope(m) = 2/3
(-4,8)...x = -4 and y = 8
sub and find b
8 = 2/3(-4) + b
8 = -8/3 + b
8 + 8/3 = b
24/3 + 8/3 = b
32/3 = b
equation is : y = 2/3x + 32/3 or 3y = 2x + 32
when y = 10
3(10) = 2x + 32
30 - 32 = 2x
-2 = 2x
-2/2 = x
-1 = x <===
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(9,0)(3,y)...slope = 1/2

y = mx + b
slope(m) = 1/2
(9,0)...x = 9 and y = 0
sub and find b
0 = 1/2(9) + b
0 = 9/2 + b
-9/2 = b
equation is : y = 1/2x - 9/2 or 2y = x - 9
sub in 3 for x and solve for y
2y = 3 - 9
2y = -6
y = -6/2
y = -3 <==
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(8,7)(11,7)...notice how the y values are the same. This means it is a horizontal line. A horizontal line has a slope of 0.
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sorry...not too good with the picture graphs