In the diagram, K(−1; 2), L and N(1; −1) are vertices of ΔKLN such that ∠LKN = 78,69º - NSC Mathematics - Question 3 - 2018 - Paper 2

The angle θ can be found using the tangent function:

$\tan(θ) = m_{KN} = -\frac{3}{2}$

To find θ, we compute:

$θ = \tan^{-1}(-\frac{3}{2})$

Solving, we find that:

$θ = 180° - 56,31° = 123,69°$.

To find the gradient of KL, we need the coordinates of points K and L. Given K(-1, 2), let's assume L is at (x, y). The gradient is given by:

$m_{KL} = \frac{y - 2}{x - (-1)} = \frac{y - 2}{x + 1}$

For KL to have a gradient of 1, set:

$\frac{y - 2}{x + 1} = 1$

This implies that:

$y - 2 = x + 1$

Thus, reorganizing gives us: $y = x + 3$.

From the previous step, we know that:

$y = x + 3$

This is the equation of the line KL in the form y = mx + c, where m = 1 and c = 3.

To calculate the length of KN, use the distance formula:

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

Substituting K(-1, 2) and N(1, -1):

$KN = \sqrt{(1 - (-1))^2 + (-1 - 2)^2} = \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$.

Let L be coordinates (x, y), given the line equation from 3.3, we have:

$y = x + 3$

We find possible coordinates by substituting feasible x-values.

Since KLNM is a parallelogram, we have:

Midpoint condition: $L = M + (K - N)$

Using the midpoint of K and N, we can solve for L's coordinates.

To calculate the area of triangle KTN, we use:

$\text{Area} = \frac{1}{2} \times b \times h$

Using base = KT and height along KN, we can compute the area as specified in the sub-steps.