Prove that the lines `y=sqrt3x+1,\ y=4\ n d\ y=-sqrt3x+2` form an equilateral triangle.

Explanation

Given: y = √3x + 1…… (1)

y = 4 ……. (2)

y = – √3x + 2……. (3)

Let us assume in triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

By solving equations (1) and (2), we get x = √3, y = 4

Thus, AB and BC intersect at B(√3,4)

Now, solving equations (1) and (3), we get x = 1/2√3, y = 3/2
Thus, AB and CA intersect at A (1/2√3, 3/2)
Similarly, solving equations (2) and (3), we get x = -2/√3, y = 4
Thus, BC and AC intersect at C (-2/√3,4)
Now, we have:

AB=5/`sqrt(3)`

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