The base BC of an equilateral triangle ABC, lies on y-axis. The coordinates of point C are (0, -3). The origin is the midpoint of the base.
Find the coordinates of the points A and B.

Explanation

To find the coordinates of points A and B for the equilateral triangle ABC, where the base BC lies on the y-axis and point C is given as (0, -3), we can follow these steps:

### Step 1: Determine the coordinates of point B
Since the origin (0, 0) is the midpoint of the base BC, and point C is at (0, -3), we can find the coordinates of point B. The y-coordinate of point B must be equal to the negative of the y-coordinate of point C because the midpoint is at the origin.

Let the coordinates of B be (0, y_B). Since the midpoint is at (0, 0), we can set up the equation:
\[ \frac{y_B + (-3)}{2} = 0 \]
Solving for \(y_B\):
\[ y_B - 3 = 0 \implies y_B = 3 \]
Thus, the coordinates of point B are (0, 3).

### Step 2: Calculate the length of side BC
Now, we can calculate the length of side BC using the distance formula:
\[ BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of points B (0, 3) and C (0, -3):
\[ BC = \sqrt{(0 - 0)^2 + (3 - (-3))^2} = \sqrt{0 + (3 + 3)^2} = \sqrt{6^2} = 6 \]

### Step 3: Determine the coordinates of point A
Since triangle ABC is equilateral, all sides are equal. Thus, the length of side AB is also 6. Let the coordinates of point A be (x_A, 0) since point A lies on the x-axis.

We can use the distance formula again for side AB:
\[ AB = \sqrt{(x_A - 0)^2 + (0 - 3)^2} \]
Setting this equal to 6:
\[ \sqrt{x_A^2 + 3^2} = 6 \]
Squaring both sides:
\[ x_A^2 + 9 = 36 \]
Solving for \(x_A^2\):
\[ x_A^2 = 36 - 9 = 27 \]
Taking the square root:
\[ x_A = \pm \sqrt{27} = \pm 3\sqrt{3} \]

### Final Coordinates
Thus, the coordinates of point A can be either:
- \(A(3\sqrt{3}, 0)\) or
- \(A(-3\sqrt{3}, 0)\)

### Summary of Coordinates
- Point B: (0, 3)
- Point C: (0, -3)
- Point A: (3√3, 0) or (-3√3, 0)