The diagram below shows a side view of a slanted ladder KL against a vertical wall KZ - NSC Technical Mathematics - Question 1 - 2021 - Paper 2

The length of KL can be determined using the distance formula:

$KL = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Where K(1; 7) and L(-3; -1) are the coordinates:

$KL = \\sqrt{(1 - (-3))^2 + (7 - (-1))^2} \\ = \\sqrt{(1 + 3)^2 + (7 + 1)^2} \\ = \\sqrt{(4)^2 + (8)^2} \\ = \\sqrt{16 + 64} \\ = \\sqrt{80} \\ = 4\, 5$

The midpoint M of segment KL can be found using the midpoint formula:

$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Substituting the coordinates of K and L:

$M = \left( \frac{1 + (-3)}{2}, \frac{7 + (-1)}{2} \right) \\ = \left( \frac{-2}{2}, \frac{6}{2} \right) \\ = (-1; 3)$

The gradient (m) of the line KL can be calculated using the formula:

$m_{KL} = \frac{y_2 - y_1}{x_2 - x_1}$

Using K(1; 7) and L(-3; -1):

$m_{KL} = \frac{-1 - 7}{-3 - 1} \\ = \frac{-8}{-4} \\ = 2$

The angle θ can be determined using the tangent function:

$\tan(\theta) = m_{KL} = 2$

Thus,

$\theta = \tan^{-1}(2) \\ \approx 63.4^\circ$

A line parallel to KL will have the same gradient, which we calculated as 2. Using the point-slope form of the line equation,

$y - y_1 = m(x - x_1)$

Substituting m = 2 and the point (-5; 1):

$y - 1 = 2(x + 5) \\ y - 1 = 2x + 10 \\ y = 2x + 11$

To determine if the point (-4; -2) lies on the line y = 2x + 11, we substitute x = -4 into the equation:

Substituting:

$y = 2(-4) + 11 \\ = -8 + 11 \\ = 3$

Since the calculated value y = 3 does not equal -2, the point (-4; -2) does NOT lie on the line.