In the diagram, P, R(3 ; 5), S(-3 ; -7) and T(-5 ; k) are vertices of trapezium PRST and PT | RS - NSC Mathematics - Question 3 - 2019 - Paper 2

To calculate the gradient (m) of RS, which connects points R(3, 5) and S(-3, -7), use the formula:

$m_{RS} = \frac{y_2 - y_1}{x_2 - x_1}$ where (x1, y1) is R and (x2, y2) is S:

$m_{RS} = \frac{-7 - 5}{-3 - 3} = \frac{-12}{-6} = 2$.

The angle θ can be calculated using the tangent function and the gradient of RS. Employing the relationship:

$tan(θ) = m_{RS}$ Substituting the value of the gradient:

$tan(θ) = 2$. Therefore, using the inverse tangent function:

$θ = \tan^{-1}(2) ≈ 63.43^ ext{o}$.

To find the coordinates of D, consider the equations of lines RS and PR. The equation of PR is:

$y = 5$. Substituting in the equation of RS to find the intersection:

$5 = -2x - 7$ Solving gives: $2x = -12 ⇒ x = -6$. Thus, coordinates D are (-6, 5).

Given that TS is the distance between points T(-5, k) and S(-3, -7):

Using the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, Set equal to given distance:

$2/\sqrt{5} = \sqrt{(-3 - (-5))^2 + (-7 - k)^2}$. Simplifying leads to:

$2/\sqrt{5} = \sqrt{4 + (k + 7)^2}$. Squaring both sides and isolating k gives:

$4/5 = 4 + (k + 7)^2$, thus:

$(k + 7)^2 = -16/5$. This results in k being -3.

For parallelogram TDNS with N in the 4th quadrant, use the midpoint theorem:

The midpoint of TN equals the midpoint of SD.

Given the coordinates, apply the midpoint formula to find N:

$N = \left(\frac{(-5 + -3)}{2}, \frac{(k + -7)}{2}\right)$. Calculating yields coordinates N.

For the reflection of APRD across the y-axis, utilize the coordinates derived:

The rotation formula around a point can also apply:

$R' = (-3, 5)$ and calculate edge angles using trigonometry, confirming the final size of angles.