Question 165289This question is from textbook Geometry
: Given: triangle ABC with coordinates A(0,0);B(a,0); and C(b,c). S is the midpoint of AC and T is the midpoint of BC.
Prove ST=1/2 AB
This question is from textbook Geometry
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! given:
point S is midpoint of AC
point T is midpoint of BC
point A = (0,0)
point C = (b,c)
point B = (a,0)
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formula for coordinates of midpoint of a line is ((x1+x2)/2,(y1+y2)/2)
let point A = (x1,y1) = (0,0)
x1 = 0
y1 = 0
let point C = (x2,y2) = (b,c)
x2 = b
y2 = c
formula for midpoint of AB is (b/2,c/2) = point S.
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formula for coordinates of midpoint of a line is ((x1+x2)/2,(y1+y2)/2)
let point B = (x1,y1) = (a,0)
x1 = a
y1 = 0
let point C = (x2,y2) = (b,c)
x2 = b
y2 = c
formula for midpoint of BC = ((a+b)/2,c/2) = point T.
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formula for length of a line is sqrt[(x2-x1)^2 + (y2-y1)^2]
let point S be (x1,y1) = (b/2,c/2)
let point T be (x2,y2) = (a+b)/2,c/2)
length of ST = sqrt[(x2-x1)^2 + (y2-y1)^2]
length of ST = sqrt [ ( (a+b)/2 - b/2 )^2 + ( c/2 - c/2 )^2 ]
length of ST = sqrt [ ( (a+b-b)/2 )^2 + 0^2 ]
length of ST = sqrt [ ( a/2 )^2 ]
length of ST = sqrt [ ( a^2/4 ) ]
length of ST = sqrt (a^2) / sqrt (4)
length of ST = a/2
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point A = (0,0)
point B = (a,0)
let point A = (x1,y1) = (0,0)
let point B = (x2,y2) = (a,0)
formula for length of a line is sqrt[(x2-x1)^2 + (y2-y1)^2]
length of line AB = sqrt [ (a-0)^2 + (0-0)^2 ]
length of line AB = sqrt [ a^2 + (0)^2 ]
length of line AB = sqrt [ a^2 ]
length of line AB = a
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length of ST = a/2 = (1/2)*a
length of AB = a
length of ST = 1/2 * length of AB
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