Shortcut Trick
For an equilateral triangle with vertices (x1, y1), (x2, y2), and (x3, y3), the y-coordinate of the centroid is Gy = (y1 + y2 + y3) ÷ 3.
In an equilateral triangle, the third vertex (x3, y3) is related to the others by symmetry. For a fixed base (0, 0) and (3, √3), the two possible third vertices (p, q1) and (p, q2) are reflections across the perpendicular bisector.
The sum of the y-coordinates of the two possible third vertices for an equilateral triangle based on points (x1, y1) and (x2, y2) is always y1 + y2.
∴ q1 + q2 = 0 + √3 = √3.
∴ The correct answer is √3.
Alternate Method
Given: Vertices A(0, 0), B(3, √3), and C(p, q).
Formula Used: Distance = √[(x2−x1)2 + (y2−y1)2]
⇒ Side AB2 = (3 − 0)2 + (√3 − 0)2 = 9 + 3 = 12.
⇒ Since it is equilateral, AC2 = 12 and BC2 = 12.
⇒ AC2 = p2 + q2 = 12 ... (i)
⇒ BC2 = (p − 3)2 + (q − √3)2 = 12.
⇒ p2 + 9 − 6p + q2 + 3 − 2√3q = 12.
⇒ (p2 + q2) + 12 − 6p − 2√3q = 12.
⇒ 12 + 12 − 6p − 2√3q = 12 ⇒ 6p + 2√3q = 12 ⇒ 3p + √3q = 6.
⇒ p = (6 − √3q) ÷ 3. Substitute this into (i):
⇒ [(6 − √3q) ÷ 3]2 + q2 = 12.
⇒ (36 + 3q2 − 12√3q) ÷ 9 + q2 = 12.
⇒ 36 + 3q2 − 12√3q + 9q2 = 108.
⇒ 12q2 − 12√3q − 72 = 0 ⇒ q2 − √3q − 6 = 0.
⇒ This is a quadratic equation in q with roots q1 and q2.
⇒ Sum of roots (q1 + q2) = −b ÷ a = −(−√3) ÷ 1 = √3.
∴ The correct answer is √3.
Additional Information
Third Vertex of Equilateral Triangle
If two vertices are (x1, y1) and (x2, y2), the third vertex (x, y) is given by x = (x1+x2 ± √3(y1−y2))/2 and y = (y1+y2 ± √3(x2−x1))/2.
Area of Equilateral Triangle
Area = (√3 ÷ 4) × side2. In this case, Area = (√3 ÷ 4) × 12 = 3√3 sq. units.
Centroid Property
For an equilateral triangle, the centroid, circumcenter, orthocenter, and incenter all coincide at the same point.