If the two vertices of an equilateral triangle are (3,0) and (6,0), find the third vertex
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let third vertex be (x+y)
all sides are equal in equilateral triangle
so,AB=BC= CA
AB = √(6-3)²+0
=3
BC= √(x-6)²+(y-0)²
9=(x-6)²+y²
CA²= (3-x)²+(0-y)²
CA² =(3-x)²+y²
so,(x-6)²+y²=(3-x)²+y²
= x²+36-12x=9+x²-6x
=36-9=6x
=27=6x
= x=27/6
all sides are equal in equilateral triangle
so,AB=BC= CA
AB = √(6-3)²+0
=3
BC= √(x-6)²+(y-0)²
9=(x-6)²+y²
CA²= (3-x)²+(0-y)²
CA² =(3-x)²+y²
so,(x-6)²+y²=(3-x)²+y²
= x²+36-12x=9+x²-6x
=36-9=6x
=27=6x
= x=27/6
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