To solve this problem, you need to find the slope of the secant line $PQ$ between the fixed point $P(7, 1)$ on the curve $y = x - 6$ and a point $Q(x, y)$ where $y = x - 6$. The slope of the secant line between points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by the formula:

$m_{\text{secant}} = \frac{y_2 - y_1}{x_2 - x_1}$

In this case, $P(7, 1)$ corresponds to $(x_1, y_1) = (7, 1)$, and $Q(x, x-6)$ corresponds to $(x_2, y_2) = (x, x-6)$. So, the slope of the secant line $PQ$ is:

$m_{\text{secant}} = \frac{(x - 6) - 1}{x - 7} = \frac{x - 7}{x - 7} = 1$

However, let's confirm this by calculating the slope for each given value of $x$:

Calculations:

1. For $x = 6.5$: $m_{\text{secant}} = \frac{(6.5 - 6) - 1}{6.5 - 7} = \frac{0.5 - 1}{-0.5} = \frac{-0.5}{-0.5} = 1.000000$

2. For $x = 6.9$: $m_{\text{secant}} = \frac{(6.9 - 6) - 1}{6.9 - 7} = \frac{0.9 - 1}{-0.1} = \frac{-0.1}{-0.1} = 1.000000$

3. For $x = 6.99$: $m_{\text{secant}} = \frac{(6.99 - 6) - 1}{6.99 - 7} = \frac{0.99 - 1}{-0.01} = \frac{-0.01}{-0.01} = 1.000000$

4. For $x = 6.999$: $m_{\text{secant}} = \frac{(6.999 - 6) - 1}{6.999 - 7} = \frac{0.999 - 1}{-0.001} = \frac{-0.001}{-0.001} = 1.000000$

5. For $x = 7.5$: $m_{\text{secant}} = \frac{(7.5 - 6) - 1}{7.5 - 7} = \frac{1.5 - 1}{0.5} = \frac{0.5}{0.5} = 1.000000$

6. For $x = 7.1$: $m_{\text{secant}} = \frac{(7.1 - 6) - 1}{7.1 - 7} = \frac{1.1 - 1}{0.1} = \frac{0.1}{0.1} = 1.000000$

7. For $x = 7.01$: $m_{\text{secant}} = \frac{(7.01 - 6) - 1}{7.01 - 7} = \frac{1.01 - 1}{0.01} = \frac{0.01}{0.01} = 1.000000$

8. For $x = 7.001$: $m_{\text{secant}} = \frac{(7.001 - 6) - 1}{7.001 - 7} = \frac{1.001 - 1}{0.001} = \frac{0.001}{0.001} = 1.000000$

Conclusion:

For all the values of $x$, the slope of the secant line $PQ$ is consistently 1.000000, correct to six decimal places.

Would you like more details on any part of this explanation? Here are some related questions to consider:

1. What is the significance of the slope of the secant line in calculus?
2. How does the slope of the secant line relate to the concept of a derivative?
3. How do we calculate the slope of a secant line if the function is more complex?
4. What happens to the slope of the secant line as $Q$ approaches $P$?
5. Can you explain the difference between a secant line and a tangent line?

Tip: When calculating slopes, always be cautious about subtracting very close numbers, as this can lead to significant rounding errors in practical calculations.