The base of an equilateral triangle with side lies along the -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.
Let the base of the equilateral triangle lie along the y-axis with midpoint at origin. Since the base has length 2a and its midpoint is at origin, the two vertices on the base are (0, a) and (0, -a).
The third vertex lies on the perpendicular bisector of the base. Since the base is on the y-axis, the perpendicular bisector is the x-axis. Let the third vertex be (x, 0).
Using distance formula, distance from (x, 0) to (0, a) equals side 2a: √[(x-0)² + (0-a)²] = 2a x² + a² = 4a² x² = 3a² x = ±a
Therefore, the vertices are (0, a), (0, -a), (a, 0) or (0, a), (0, -a), (-a, 0).
Explanation
This question tests understanding of coordinate geometry with equilateral triangles. The key insight is that the perpendicular bisector of the base passes through the third vertex. Since the base lies on the y-axis with midpoint at origin, the perpendicular bisector is the x-axis. The distance formula is then used to find the x-coordinate of the third vertex by equating it to the side length 2a.
Solution Steps
Step 1: Identify base vertices as (0, a) and (0, -a) since midpoint is origin and base length is 2a.
Step 2: Recognize that third vertex lies on x-axis (perpendicular bisector of base).
Step 3: Apply distance formula between (x, 0) and (0, a) equal to 2a.
Step 4: Solve x² + a² = 4a² to get x = ±a.
Step 5: State both possible sets of vertices.