To determine whether a set of three points lies on a straight line, we can use the concept of slopes. The points will be collinear if the slope between any two pairs of points is the same.

Formula for slope:

$\text{slope (m)} = \frac{y_2 - y_1}{x_2 - x_1}$

We'll calculate the slopes for each set of points, comparing the slopes between:

1. $A$ and $B$
2. $B$ and $C$
3. $A$ and $C$

a) $A(-1, 2)$, $B(2, 1)$, $C(5, 0)$

1. Slope between $A(-1, 2)$ and $B(2, 1)$: $m_{AB} = \frac{1 - 2}{2 - (-1)} = \frac{-1}{3} = -\frac{1}{3}$

2. Slope between $B(2, 1)$ and ( C(5, 0): $m_{BC} = \frac{0 - 1}{5 - 2} = \frac{-1}{3} = -\frac{1}{3}$

3. Slope between $A(-1, 2)$ and ( C(5, 0): $m_{AC} = \frac{0 - 2}{5 - (-1)} = \frac{-2}{6} = -\frac{1}{3}$

Since $m_{AB} = m_{BC} = m_{AC} = -\frac{1}{3}$, the points $A(-1, 2)$, $B(2, 1)$, and $C(5, 0)$ are collinear.

b) $A(-10, 2)$, $B(1, -2)$, $C(6, -5)$

1. Slope between $A(-10, 2)$ and ( B(1, -2): $m_{AB} = \frac{-2 - 2}{1 - (-10)} = \frac{-4}{11} = -\frac{4}{11}$

2. Slope between $B(1, -2)$ and ( C(6, -5): $m_{BC} = \frac{-5 - (-2)}{6 - 1} = \frac{-3}{5} = -\frac{3}{5}$

3. Slope between $A(-10, 2)$ and ( C(6, -5): $m_{AC} = \frac{-5 - 2}{6 - (-10)} = \frac{-7}{16}$

Since $m_{AB} \neq m_{BC} \neq m_{AC}$, the points $A(-10, 2)$, $B(1, -2)$, and $C(6, -5)$ are not collinear.

Conclusion:

• Set a) $A(-1, 2)$, $B(2, 1)$, and $C(5, 0)$ are collinear.
• Set b) $A(-10, 2)$, $B(1, -2)$, and $C(6, -5)$ are not collinear.

Would you like further details or have any other questions?

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Relative questions:

1. What is the general condition for three points to be collinear?
2. How can the slope formula be derived geometrically?
3. How do parallel lines relate to slopes?
4. How does changing the position of one point affect collinearity?
5. Can three-dimensional points be checked for collinearity similarly?

Tip: When checking for collinearity, ensuring the slopes between different pairs of points are identical is crucial!