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.