To show that the points $A(3, 9)$, $B(-2, -16)$, and $C\left(\frac{1}{5}, 5\right)$ are collinear, we need to check if the slope between any two pairs of points is the same. If all the slopes between the points are equal, then the points lie on the same straight line, hence they are collinear.

Step 1: Calculate the slope between $A$ and $B$

The formula for the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

For points $A(3, 9)$ and $B(-2, -16)$:

$m_{AB} = \frac{-16 - 9}{-2 - 3} = \frac{-25}{-5} = 5$

Step 2: Calculate the slope between $B$ and $C$

For points $B(-2, -16)$ and $C\left(\frac{1}{5}, 5\right)$:

$m_{BC} = \frac{5 - (-16)}{\frac{1}{5} - (-2)} = \frac{5 + 16}{\frac{1}{5} + 2} = \frac{21}{\frac{11}{5}} = 21 \times \frac{5}{11} = \frac{105}{11}$

Step 3: Calculate the slope between $A$ and $C$

For points $A(3, 9)$ and $C\left(\frac{1}{5}, 5\right)$:

$m_{AC} = \frac{5 - 9}{\frac{1}{5} - 3} = \frac{-4}{\frac{1}{5} - \frac{15}{5}} = \frac{-4}{-\frac{14}{5}} = -4 \times \frac{5}{-14} = \frac{20}{14} = \frac{10}{7}$

Conclusion:

Since the slopes $m_{AB} = 5$, $m_{BC} = \frac{105}{11}$, and $m_{AC} = \frac{10}{7}$ are not equal, the points are not collinear.

Please feel free to ask any questions if you'd like more details, or if something is unclear!

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Here are 5 related questions for deeper understanding:

1. How can you find the equation of a line through two points?
2. What is the geometrical significance of collinear points?
3. How can you generalize the slope formula to 3D points?
4. What does the determinant test for collinearity in vectors involve?
5. Can you apply the midpoint formula between two of these points?

Tip: The slopes between three collinear points must always be equal for them to lie on a straight line.