Find the points on the line y=3-x which are 2 units from (1,1)

We are tasked with finding the points on the line \( y = 3 - x \) that are 2 units away from the point \( (1, 1) \). To do this, we can use the distance formula and solve for the points on the line.

### Step-by-step approach:

1. **The equation of the line**: The points on the line \( y = 3 - x \) can be written as \( (x, 3 - x) \), where \( x \) is a variable.

2. **Distance formula**: The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

   \[
   d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
   \]

   In our case, the point \( (1, 1) \) is fixed, and we are finding points \( (x, 3 - x) \) on the line such that the distance between \( (1, 1) \) and \( (x, 3 - x) \) is 2 units. Thus, we set up the distance equation:

   \[
   \sqrt{(x - 1)^2 + \left( (3 - x) - 1 \right)^2} = 2
   \]

3. **Simplify the equation**:
   
   - \( (3 - x) - 1 = 2 - x \)
   
   Substituting this into the equation:

   \[
   \sqrt{(x - 1)^2 + (2 - x)^2} = 2
   \]

4. **Square both sides** to eliminate the square root:

   \[
   (x - 1)^2 + (2 - x)^2 = 4
   \]

5. **Expand both terms**:
   
   - \( (x - 1)^2 = x^2 - 2x + 1 \)
   - \( (2 - x)^2 = (x - 2)^2 = x^2 - 4x + 4 \)
   
   Now substitute these expansions into the equation:

   \[
   (x^2 - 2x + 1) + (x^2 - 4x + 4) = 4
   \]

6. **Combine like terms**:

   \[
   2x^2 - 6x + 5 = 4
   \]

   Simplifying:

   \[
   2x^2 - 6x + 1 = 0
   \]

7. **Solve the quadratic equation**:
   
   We can solve the quadratic equation \( 2x^2 - 6x + 1 = 0 \) using the quadratic formula:

   \[
   x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(1)}}{2(2)}
   \]

   Simplifying:

   \[
   x = \frac{6 \pm \sqrt{36 - 8}}{4} = \frac{6 \pm \sqrt{28}}{4} = \frac{6 \pm 2\sqrt{7}}{4}
   \]

   \[
   x = \frac{3 \pm \sqrt{7}}{2}
   \]

8. **Find the corresponding \( y \)-values**:
   
   Since \( y = 3 - x \), substitute the values of \( x \) into the line equation:
   
   - For \( x = \frac{3 + \sqrt{7}}{2} \), \( y = 3 - \frac{3 + \sqrt{7}}{2} = \frac{3 - \sqrt{7}}{2} \)
   - For \( x = \frac{3 - \sqrt{7}}{2} \), \( y = 3 - \frac{3 - \sqrt{7}}{2} = \frac{3 + \sqrt{7}}{2} \)

### Final answer:
The points on the line \( y = 3 - x \) that are 2 units away from \( (1, 1) \) are:

\[
\left( \frac{3 + \sqrt{7}}{2}, \frac{3 - \sqrt{7}}{2} \right) \quad \text{and} \quad \left( \frac{3 - \sqrt{7}}{2}, \frac{3 + \sqrt{7}}{2} \right)
\]

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Would you like more details on any of the steps?

Here are 5 related questions to expand on this concept:

1. How can we verify the distance between two points using the distance formula?
2. How does the quadratic formula work in solving quadratic equations?
3. What is the geometric interpretation of points equidistant from a given point and a line?
4. Can the distance formula be generalized to 3-dimensional space?
5. How do we apply transformations to lines and circles to simplify distance calculations?

**Tip:** Always double-check your algebraic simplifications, especially when solving quadratic equations!

Learn how to find points on the line y=3-x that are exactly 2 units away from the point (1,1). This problem involves using the distance formula, solving a quadratic equation, and applying algebraic simplifications. Perfect for students in Grades 9-11.

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