Question

If 'a' is the length of one of the sides of an equilateral triangle ABC, base BC lies on x-axis and vertex B is at origin, find the coordinates of the vertices of the triangle ABC.

Answer 1

Coordinates of the vertex B = (0, 0)
Coordinates of the vertex C = (a, 0) [ Since the length of the side is a units]
Let the coordianates of the vertex A be (x, y)
AB = x 2 + y 2 =a ------------- (1)
AC = (x-a) 2 + y2 =a ------------- (2)

AB = AC

x 2 + y 2 = (x-a) 2 + y 2
x2 + y2 = x2 + a2 - 2ax + y2
a2 - 2ax = 0
a(a - 2x) = 0
a - 2x = 0
a = 2x
x = a/2
From equation (1)
x2 + y2 = a2
a2/4 + y2 = a2
y2 = a2 – a2/4
y2 = 3/4 a2
y = √[3/4 a2]
y = (√3/2) a

Hence, the coordinates of the vertex A are [a/2, (√3/2) a]. Coordiantes of B are (0, 0).Coordinates of C are (a, 0).