In the diagram, the equation of line AF is $y = -x - 11$ - NSC Mathematics - Question 3 - 2023 - Paper 2

We know that the tangent of the angle $\alpha$ formed between the line and the positive x-axis can be derived from its slope. For the line $y = -x - 11$, the slope (m) is -1. Thus:

$\tan(\alpha) = \frac{\text{rise}}{\text{run}} = -1.$ This corresponds to an angle of:

$\alpha = 135^{\circ}.$

For the angle $63.43^{\circ}$, we calculate the gradient of line AC:

$\tan(63.43^{\circ}) = m_{AC} \approx 2.$ Thus, the gradient of AC to the nearest whole number is 2.

Given that the gradient $m_{AC} = 2$ and passing through point A(-1, -10), we can find the equation of line AC. Using the point-slope form:

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

Substituting:

$y - (-10) = 2(x - (-1))$

Simplifying, we get:

$y + 10 = 2(x + 1) \\ y = 2x - 8.$

To find the coordinates of point C, we rewrite line AC using the x-axis intersection:

Let $x_C = 4$ (as given at point B), then:

$y_{C} = 2(4) - 8 = 0.$ Thus, Coordinates of C are (4; 0).

The size of angle $F\overset{~}{E}D$ can be determined using corresponding angles, where:

$\angle ABE = 63.43^{\circ} \\ \therefore \angle F\overset{~}{E}D = 180^{\circ} - 63.43^{\circ} = 116.57^{\circ}.$

To derive the equation of the circle centered at G and passing through point B(4, 10), we must first determine G's coordinates. Assuming G is at G(-11, 19) and using the standard circle equation:

$(x - a)^{2} + (y - b)^{2} = r^{2},$

we calculate:

$r^2 = (4 + 11)^{2} + (10 - 19)^{2} = 15^{2} + (-9)^{2} = 225 + 81 = 306.$ Thus, the equation is:

$(x + 11)^{2} + (y - 19)^{2} = 306.$