The base QR of an equilateral triangle PQR lies on X-axis. The co-ordinates of the point Q are (-4,0) and origin is the mid-point of the base. Find the co-ordinates of the points P and R.

Answer

To solve the problem, we will follow these steps:

1. **Identify the coordinates of points Q and R**:
- Given that Q has coordinates (-4, 0) and the origin (0, 0) is the midpoint of the base QR, we can find the coordinates of R.
- Since the origin is the midpoint, the distance from Q to the origin is equal to the distance from the origin to R.

2. **Calculate the coordinates of point R**:
- The distance from Q (-4, 0) to the origin (0, 0) is 4 units.
- Therefore, point R must be 4 units to the right of the origin, which gives us the coordinates of R as (4, 0).

3. **Determine the coordinates of point P**:
- Since P is the apex of the equilateral triangle and lies above the x-axis, we denote its coordinates as (0, y).
- The lengths of the sides PQ, QR, and PR must all be equal.

4. **Use the distance formula to find y**:
- The distance PQ can be calculated as:
\[ PQ = \sqrt{(0 - (-4))^2 + (y - 0)^2} = \sqrt{(4)^2 + y^2} = \sqrt{16 + y^2} \]
- The distance QR is:
\[ QR = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{(8)^2} = 8 \]
- The distance PR can be calculated as:
\[ PR = \sqrt{(0 - 4)^2 + (y - 0)^2} = \sqrt{(-4)^2 + y^2} = \sqrt{16 + y^2} \]

5. **Set the distances equal**:
- Since PQ = QR, we have:
\[ \sqrt{16 + y^2} = 8 \]
- Squaring both sides gives:
\[ 16 + y^2 = 64 \]
- Rearranging gives:
\[ y^2 = 64 - 16 = 48 \]
- Thus, \( y = \pm \sqrt{48} = \pm 4\sqrt{3} \).

6. **Final coordinates of point P**:
- Therefore, the coordinates of point P can be (0, 4√3) or (0, -4√3). Since P is above the x-axis, we take (0, 4√3).

### Summary of Coordinates:
- Q = (-4, 0)
- R = (4, 0)
- P = (0, 4√3)