Question 26 in the image reads:

"26. Prove that the following sets of points are collinear:

(i) $(1, 1), (3, 2), (5, 3)$

(ii) $(1, -3), (4, -6), (8, -16)$

(iii) $(a + b, c), (b, c + a), (c, a + b)$"

Let's solve these parts one by one by checking if the points lie on the same line, i.e., if the slopes between any two points are equal.

Part (i) – Prove that $(1, 1)$, $(3, 2)$, and $(5, 3)$ are collinear:

The slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

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

1. Slope between $(1, 1)$ and $(3, 2)$:

$m_{12} = \frac{2 - 1}{3 - 1} = \frac{1}{2}$

2. Slope between $(3, 2)$ and $(5, 3)$:

$m_{23} = \frac{3 - 2}{5 - 3} = \frac{1}{2}$

3. Slope between $(1, 1)$ and $(5, 3)$:

$m_{13} = \frac{3 - 1}{5 - 1} = \frac{2}{4} = \frac{1}{2}$

Since the slopes between all pairs of points are equal, the points $(1, 1)$, $(3, 2)$, and $(5, 3)$ are collinear.

Part (ii) – Prove that $(1, -3)$, $(4, -6)$, and $(8, -16)$ are collinear:

1. Slope between $(1, -3)$ and $(4, -6)$:

$m_{12} = \frac{-6 - (-3)}{4 - 1} = \frac{-6 + 3}{3} = \frac{-3}{3} = -1$

2. Slope between $(4, -6)$ and $(8, -16)$:

$m_{23} = \frac{-16 - (-6)}{8 - 4} = \frac{-16 + 6}{4} = \frac{-10}{4} = -2.5$

Since the slopes between these points are not equal ($m_{12} \neq m_{23}$), the points $(1, -3)$, $(4, -6)$, and $(8, -16)$ are not collinear.

Part (iii) – Prove that $(a + b, c)$, $(b, c + a)$, and $(c, a + b)$ are collinear:

The points are expressed in terms of variables. To prove collinearity, we calculate the determinant of the matrix formed by the coordinates of these points. If the determinant is 0, the points are collinear.

The determinant is: