A line that passes through the origin intersects both the line x = 1

Let the line that passes thru origin be line L. This line intersects two lines: X=1 and Y=\(1+\sqrt{3}/3\) x
Note that the line Y=\(1+\sqrt{3}/3\) x makes 30 deg angle with horizontal since the slope is \(1/\sqrt{3}\)
This makes it impossible for L to intersect the two lines in the first quadrant

The three lines will intersect in a manner shown in the image. We are required to find the perimeter of equilateral triangle ABC.
The line Y=\(1+\sqrt{3}/3\) x intersects Y axis at 1 (point D)
(The origin of coordinate system is at O)

Now consider right triangle POB:
Angle PBO=60 and angle BPO=90. So, angle BOP=30. This makes it a 30-60-90 triangle where ratio PB:PO:BO will be \(1:\sqrt{3}:2\).
We know that PO=1. Therefore PB=\(1/\sqrt{3}\), BO=\(2/\sqrt{3}\).

Now consider triangle COD:
Angle OCD=60. Also, angle DOC and CDO=60 and hence triangle COD is also an equilateral triangle.
We know that DO=1. Since COD is an equilateral triangle, DO=CD=CO=1.

We now already know the length of side CB of equilateral triangle ABC
CB=CO+BO=1+\(2/\sqrt{3}\)

As the triangle is equilateral, AB=BC=AC.
Perimeter= 3*(1+\(2/\sqrt{3})\)= 3+2\(\sqrt{3}\)

Answer: D

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