get all coordinates of P on x axis so that line that cross A(1,2) and P are perpendicular to line that cross B(8,3) and P

To find all coordinates \( P(x, 0) \) on the \( x \)-axis such that the line passing through \( A(1, 2) \) and \( P \) is perpendicular to the line passing through \( B(8, 3) \) and \( P \), we need to use the condition for perpendicular lines.

### Step-by-Step Solution:

1. **Equation for the slopes**: The condition for perpendicular lines is that the product of their slopes must be \( -1 \). If the slope of the line passing through points \( A \) and \( P \) is \( m_1 \), and the slope of the line passing through points \( B \) and \( P \) is \( m_2 \), we have:
   \[
   m_1 \cdot m_2 = -1
   \]

2. **Slope of line \( AP \)**:
   The slope \( m_1 \) of the line through \( A(1, 2) \) and \( P(x, 0) \) is:
   \[
   m_1 = \frac{0 - 2}{x - 1} = \frac{-2}{x - 1}
   \]

3. **Slope of line \( BP \)**:
   The slope \( m_2 \) of the line through \( B(8, 3) \) and \( P(x, 0) \) is:
   \[
   m_2 = \frac{0 - 3}{x - 8} = \frac{-3}{x - 8}
   \]

4. **Perpendicularity condition**:
   Using the condition \( m_1 \cdot m_2 = -1 \), we substitute the slopes:
   \[
   \left( \frac{-2}{x - 1} \right) \cdot \left( \frac{-3}{x - 8} \right) = -1
   \]
   Simplifying this:
   \[
   \frac{6}{(x - 1)(x - 8)} = -1
   \]

5. **Solve the equation**:
   Multiply both sides by \( (x - 1)(x - 8) \):
   \[
   6 = -(x - 1)(x - 8)
   \]
   Expand the right-hand side:
   \[
   6 = -(x^2 - 9x + 8)
   \]
   Simplifying:
   \[
   6 = -x^2 + 9x - 8
   \]
   Bring everything to one side:
   \[
   x^2 - 9x + 14 = 0
   \]

6. **Solve the quadratic equation**:
   We can solve the quadratic equation \( x^2 - 9x + 14 = 0 \) using the quadratic formula:
   \[
   x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(14)}}{2(1)}
   \]
   Simplifying:
   \[
   x = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2}
   \]
   \[
   x = \frac{9 \pm 5}{2}
   \]
   So, we have two solutions:
   \[
   x = \frac{9 + 5}{2} = 7 \quad \text{or} \quad x = \frac{9 - 5}{2} = 2
   \]

### Conclusion:
The two possible coordinates for \( P \) on the \( x \)-axis are \( P(7, 0) \) and \( P(2, 0) \).

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Do you want more details on any step or have any questions?

Here are 5 related questions to consider:
1. How would the solution change if \( P \) were not constrained to the \( x \)-axis?
2. What is the geometric interpretation of the perpendicularity condition?
3. How can the condition for perpendicularity be generalized for three-dimensional space?
4. How can the equation of a perpendicular bisector be derived from two points?
5. How do you solve the quadratic equation geometrically?

**Tip**: Always check the geometry of a problem involving slopes to ensure the condition, such as perpendicularity, holds geometrically.

Get all coordinates of P on the x-axis so that the line that crosses A(1, 2) and P is perpendicular to the line that crosses B(8, 3) and P.

Discover how to find the coordinates of point P on the x-axis such that the lines passing through A(1, 2) and B(8, 3) are perpendicular. This problem involves applying the perpendicularity condition for slopes and solving a quadratic equation. Suitable for students in Grades 9-12.

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