The base AB of an equilateral triangle ABC with side 8 cm lies along the y-axis such that the mid-point of AB is at the origin and B lies above the origin. What is the equation of line passing through (8,0) and parallel to the side AC?

Explanation

Since AB=8 and the origin is the midpoint, A = (0, -4) and B = (0, 4). The third vertex C lies on the x-axis, so C = (4\sqrt{3}, 0) or (-4\sqrt{3}, 0). Taking C(4\sqrt{3}, 0) (standard orientation in the positive x-axis), the slope of AC is \frac{0 - (-4)}{4\sqrt{3} - 0} = \frac{1}{\sqrt{3}}. The equation of a line parallel to AC through (8, 0) is y - 0 = \frac{1}{\sqrt{3}}(x - 8) \implies x - \sqrt{3}y - 8 = 0.

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