SOLUTION: SHOW in two ways that (0,-5) (3,-4) (8,0) and (5,-1) are the vertices of a rhombus. PLEASSSE HELP!
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Question 864720: SHOW in two ways that (0,-5) (3,-4) (8,0) and (5,-1) are the vertices of a rhombus. PLEASSSE HELP!
Found 2 solutions by stanbon, rothauserc:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
SHOW in two ways that (0,-5) (3,-4) (8,0) and (5,-1) are the vertices of a rhombus. PLEASSSE HELP!
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1st: Plot the points so you can see where the sides would be.
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2nd: Find the length of the "sides" using:
d = sqrt[(x1-x2)^2+(y1-y2)^2]
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3rd: If the sides are equal, the figure is a rhombus.
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Example::
d = sqrt[(0-5)^2+(-5--1)^2] = sqrt[25+16) = sqrt(41)
d = sqrt((3-8)^2 + (-4-0)^2] = sqrt[25+16] = sqrt(41)
etc.
Cheers,
Stan H.
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Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
From the definition of a Rhombus, we know that all four sides are of equal length ,the opposite sides are parallel and the diagonals are of equal length.
we will use the following math concepts
distance (d) between two points = square root ( (x2-x1)^2 + (y2-y1)^2 )
two lines are parallel if they have the same slope (m)
m is defined as (y2 - y1) / (x2 - x1)
we are given the vertices (0,-5) (3,-4) (8,0) and (5,-1) and label them A,B,C,D
d(AB) = square root ( 3^2 + 1^2 ) = square root (10)
d(CD) = square root ( -3^2 + -1^2) = square root (10)
d(BC) = square root ( 5^2 + 4^2) = square root (41)
d(AD) = square root ( 5^2 + 4^2) = square root (41)
d(AC) = square root ( 8^2 + 5^2) = square root (89)
d(BD) = square root ( 2^2 + 3^2) = square root (13)
m(AB) = 1/3
m(CD) = -1/-3 = 1/3
m(BC) = 4/5
m(AD) = 4/5
m(AC) = 5/8
m(BD) = 3/2
AB and CD are parallel and equal length
BC and AD are parallel and equal length
AC and BD are not parallel or perpendicular and are of unequal length
this is NOT a rhombus
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