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Question 1207805: An equilateral triangle is one in which all three sides are of equal length. If two vertices of an equilateral triangle are (0,4) and (0,0) find the third vertex. How many of these triangles are possible?
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! Let the vertices of the equilateral triangle be A(0,4), B(0,0), and C(x,y).
The distances between all pairs of vertices must be equal.
The length ofAB=sqrt((0−0)^2+(4−0)^2)=4
AB^2=16
The length of AC = sqrt((0-x)^2+(4-y)^2))
AC^2 = x^2+16-8y+y^2
BC=sqrt((0-x)^2+(0-y)^2))
BC^2 =x^2+y^2
AC^2=BC^2 ( equilqteral triangle)
x^2+y^2=x^2+16-8y+y^2
rearrange
8y=16
y=2
substitute y = 2 in x^2+y^2=16
x^2+4=16
x^2= 12
x = +/-2sqrt(3)
C =(2sqrt(3),2) or (-2sqrt(3),2)
There are two possible equilateral triangles with the C vertex being either (2sqrt(3),2)(-2sqrt(3),2) with other two vertices being A(0,4), B(0,0),
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