In die diagram is A(3 ; 4), B en C hoeke punte van ΔABC - NSC Mathematics - Question 3 - 2024 - Paper 2

To find the x-intercept of line AS, we set y = 0 in the line equation. The equation of line AS is obtained from the coordinates of points A and S. Calculating the gradient:

$m_{AS} = \frac{y_S - 4}{x_S - 3}$

Let point S be (x_S, y_S). Setting y = 0:

$0 = \frac{0 - 4}{x_S - 3} + 4$

We solve for x_S now:

Point C can be found by substituting back into the equation of line AC, which is derived from the coordinates and gradients we have thus far. We have line AC:

$y = 2x - 2$

Substituting an appropriate x-value yields:

The slope for line BC can be found using its coordinates. Once we find the slope, we apply the point-slope form using S(-15, -2):

$y + 2 = m_{BC}(x + 15)$

Rearranging gives us the required equation:

To find the angle ∠BAC, we need to use the tangent of the slope found in line BC:

$tan(\alpha) = m_{BC} = \frac{y_B - y_A}{x_B - x_A}$

Evaluating this gives the angle using the arctangent function.

The area of triangle ΔABD can be calculated using the formula:

$Area = \frac{1}{2} \times base \times height$

Finding lengths using the coordinates derived earlier will allow us to finalize both areas.