if two vertices of a equilateral triangle are (3,0) and (6,0) find the third vertex
Anonymous:
anything about centroid mention there??
that's why I'm asking
I don't understand
its quadrilateral or equilateral??plz check
becoz i have not seen anything like quadrilateral triangle
may be its quilateral traingle
equilateral*
whose all sides are equal
sorry its equilateral
my bad
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We all know the sides of a equilateral triangle are equal.
So, in triangle ABC AB=BC=CA
we know co-ordinates of 2 points A(3,0),B(6,0) Let C be (a,b)
AB=BC=CA
ALL THE SIDES ARE EQUAL.USING DISTANCE FORMULA WE WILL CALCULATE THE LENGTH OF AB,BC,CA...
Distance between 2 points is given as √(Χ1-x2)²-(y1-y2)²
AB=√{(6-3)²+0}= 3 ...................1
BC=√((a-6)²+b²)........................2
CA=√((a-3)²+b²)........................3
AB=BC=CA,so we will equate them
AB=BC
3=√((a-6)²+b²)
9=a²+36+b²...........................4
BC=CA
(a-6)²+b²=(a-3)²+b².
a²+36-12a=a²-6a+9
on solving we get
a=9/2
put the value in eqn 4
u will get b=3√3/2
Hence third vertex of equilateral triangle=C=(9/2,3√3/2)
So, in triangle ABC AB=BC=CA
we know co-ordinates of 2 points A(3,0),B(6,0) Let C be (a,b)
AB=BC=CA
ALL THE SIDES ARE EQUAL.USING DISTANCE FORMULA WE WILL CALCULATE THE LENGTH OF AB,BC,CA...
Distance between 2 points is given as √(Χ1-x2)²-(y1-y2)²
AB=√{(6-3)²+0}= 3 ...................1
BC=√((a-6)²+b²)........................2
CA=√((a-3)²+b²)........................3
AB=BC=CA,so we will equate them
AB=BC
3=√((a-6)²+b²)
9=a²+36+b²...........................4
BC=CA
(a-6)²+b²=(a-3)²+b².
a²+36-12a=a²-6a+9
on solving we get
a=9/2
put the value in eqn 4
u will get b=3√3/2
Hence third vertex of equilateral triangle=C=(9/2,3√3/2)
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