The points A and B are shown on the co-ordinate grid below - Junior Cycle Mathematics - Question 6 - 2015

To find the length of the segment [AB], we apply the distance formula:

$|AB| = ext{sqrt}((x_2 - x_1)^2 + (y_2 - y_1)^2)$

Here, point A is (1, 4) and point B is (6, 2).

Plugging in the coordinates gives:

$|AB| = ext{sqrt}((6 - 1)^2 + (2 - 4)^2)$
$|AB| = ext{sqrt}((5)^2 + (-2)^2)$
$|AB| = ext{sqrt}(25 + 4)$
$|AB| = ext{sqrt}(29)$

Thus, the length of [AB] is:

$|AB| = ext{sqrt}(29)$

Point C is located at (–4, 1). It should be marked and labeled clearly on the co-ordinate grid.

To find the slope of line CA, we use the slope formula:

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

Here, point C is (–4, 1) and point A is (1, 4).

Thus,

$m = \frac{4 - 1}{1 - (-4)}$

Calculating the rise and run gives:

$m = \frac{3}{1 + 4} = \frac{3}{5}$

Therefore, the slope of line CA is:

$m = \frac{3}{5}$