Find the co-ordinates of a point whose absicissa is 10 and its distance from the point (2, -3) is 10 units.

Answer

To find the coordinates of a point whose abscissa is 10 and whose distance from the point (2, -3) is 10 units, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Point**:
Let the point be \( P(10, k) \), where \( k \) is the y-coordinate we need to find.

2. **Use the Distance Formula**:
The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Here, we know the distance \( d = 10 \) and the coordinates of the other point are \( (2, -3) \). Thus, we can set up the equation:
\[ \sqrt{(10 - 2)^2 + (k - (-3))^2} = 10 \]

3. **Simplify the Equation**:
Substitute the values into the equation:
\[ \sqrt{(8)^2 + (k + 3)^2} = 10 \]
This simplifies to:
\[ \sqrt{64 + (k + 3)^2} = 10 \]

4. **Square Both Sides**:
To eliminate the square root, square both sides:
\[ 64 + (k + 3)^2 = 100 \]

5. **Rearrange the Equation**:
Subtract 64 from both sides:
\[ (k + 3)^2 = 36 \]

6. **Take the Square Root**:
Taking the square root of both sides gives us two equations:
\[ k + 3 = 6 \quad \text{or} \quad k + 3 = -6 \]

7. **Solve for k**:
- From \( k + 3 = 6 \):
\[ k = 6 - 3 = 3 \]
- From \( k + 3 = -6 \):
\[ k = -6 - 3 = -9 \]

8. **Final Coordinates**:
Thus, the two possible coordinates of point \( P \) are:
\[ (10, 3) \quad \text{and} \quad (10, -9) \]

### Final Answer:
The coordinates of the point are \( (10, 3) \) and \( (10, -9) \).