Question

The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Answer 1

Key points to solve the problem: 

• Idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

• Equilateral triangle- triangle with all 3 sides equal. 

• Coordinates of midpoint of a line segment – Let P(x₁,y₁) and Q(x₂,y₂) be the end points of line segment PQ. Then coordinated of midpoint of PQ is given by – \(\big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\big)\) 

Given, an equilateral triangle with base along y axis and midpoint at (0,0)

∴ coordinates of triangle will be A(0,y₁) B(0,y₂) and C(x,0)

As midpoint is at origin ⇒ y₁+y₂ = 0 ⇒ y₁ = -y₂ …..eqn 1

Also length of each side = 2a (given)

∴ AB = \(\sqrt{(0-0)^2 + (y_2 - y_1)^2}\) = y₂ - y₁ = 2a….eqn 2 

∴ from eqn 1 and 2: 

y₁ = a and y₂ = -a 

∴ 2 coordinates are – A(0,a) and B(0,-a) 

See the figure:

Clearly from figure: 

DC = x 

Also in ΔADC: cos 30° = \(\frac{DC}{AC}\) = \(\frac{x}{\sqrt{(0-x)^2 + (a-0)^2}}\)

\(\therefore\) \(\frac{\sqrt{3}}{2}\) = \(\frac{x}{\sqrt{x^2 + a^2}}\)

Squaring both sides:

3(x² + a²) = 4x² \(\Rightarrow\) x² = 3a²

^{\(\therefore\)}  x = ±\(\sqrt{3a}\)

∴ Coordinates of C are (√3a,0) or (-√3a,0)