To determine if three points lie on a straight line, we can calculate the slopes between pairs of points. If the slopes between all pairs are the same, the points lie on the same straight line.

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

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

2. Slope between B and C: $m_{BC} = \frac{y_C - y_B}{x_C - x_B} = \frac{0 - 1}{5 - 2} = \frac{-1}{3} = -\frac{1}{3}$

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

Since $m_{AB} = m_{BC} = m_{AC}$, the points $A(-1, 2), B(2, 1), C(5, 0)$ lie on the same straight line.

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

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

2. Slope between B and C: $m_{BC} = \frac{y_C - y_B}{x_C - x_B} = \frac{5 - (-2)}{6 - 1} = \frac{7}{5}$

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

Since $m_{AB} \neq m_{BC} \neq m_{AC}$, the points $A(-10, 2), B(1, -2), C(6, 5)$ do not lie on the same straight line.

Conclusion:

• Set a): The points lie on the same line.
• Set b): The points do not lie on the same line.

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you find the equation of a line using two points?
2. What is the significance of the slope being constant between points?
3. Can you check if points lie on a line by using the equation of a line?
4. How do you find the midpoint between two points on a coordinate plane?
5. What are the properties of a line with a slope of zero?

Tip: If three points are collinear, the area of the triangle formed by them will be zero.