Prove that (2a,4a), (2a,6a) and (2a+a root3,5a) are vertices of an equilateral triangle.
just show that the sides have the same length.
For instance, the first two points define a side of length d, where
d^2 = (2a-2a)^2+(6a-4a)^2 = 4a^2
so, d=2a
Now just show that the other two sides are the same length.
We could "shrink" the figure, after all sides of all equilateral triangles are in the same ratio, so let's use
A(2,4), B(2,6), and C(2+√3 , 5)
are the angles equal to 60° each ??
clearly AB is a vertical line.
slope CA = 1/(2+√3 - 2) = 1/√3
so CA makes an angle of 60° with the x-axis
slope CB = -1/√3, so CB makes an angle of 120°
So , look at your sketch, what do your think ?
Btw, we could have found the slopes using the original points containing the "a" 's, they would have cancelled .
To prove that the points (2a, 4a), (2a, 6a), and (2a + a√3, 5a) are vertices of an equilateral triangle, we need to show that the lengths of all three sides are equal.
Let's find the distance between the points (2a, 4a) and (2a, 6a) first. The formula to find the distance between two points (x1, y1) and (x2, y2) is given by the distance formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Using this formula, we can calculate the distance:
Distance = √((2a - 2a)^2 + (6a - 4a)^2)
= √(0^2 + 2a^2)
= √(2a^2)
= a√2
So, the distance between (2a, 4a) and (2a, 6a) is a√2.
Next, let's find the distance between the points (2a, 4a) and (2a + a√3, 5a):
Distance = √((2a + a√3 - 2a)^2 + (5a - 4a)^2)
= √((a√3)^2 + a^2)
= √(3a^2 + a^2)
= √(4a^2)
= 2a
Finally, let's find the distance between (2a + a√3, 5a) and (2a, 6a):
Distance = √((2a - (2a + a√3))^2 + (6a - 5a)^2)
= √((-a√3)^2 + a^2)
= √(3a^2 + a^2)
= √(4a^2)
= 2a
Now we can see that all three sides have equal lengths: a√2, 2a, and 2a. Therefore, the points (2a, 4a), (2a, 6a), and (2a + a√3, 5a) are vertices of an equilateral triangle.