In the diagram below, PQRS is a quadrilateral with vertices P(-2 ; 2), Q(0 ; -4), R(6 ; -2) and S(4 ; m) - NSC Technical Mathematics - Question 1 - 2022 - Paper 2

The angle of inclination (θ) can be calculated using: $\tan(\theta) = m_{PR}$ Given that $m_{PR} = -\frac{1}{2}$, $\theta = \tan^{-1}(-\frac{1}{2})$. The reference angle is approximately 26.57°, thus $\theta = 180° - 26.57° = 153.43°$.

The length of QR can be found using the distance formula: $QR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Substituting the coordinates Q(0, -4) and R(6, -2), we get: $QR = \sqrt{(6 - 0)^2 + (-2 - (-4))^2} = \sqrt{(6)^2 + (2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$

The coordinates of midpoint E can be calculated using: $E = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ For points P(-2, 2) and Q(0, -4), we find: $E = \left( \frac{-2 + 0}{2}, \frac{2 + (-4)}{2} \right) = \left( -1, -1 \right)$

Since SR is parallel to PQ, they share the same gradient. The gradient of PQ is: $m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1}$ For points P(-2, 2) and Q(0, -4), we get: $m_{PQ} = \frac{-4 - 2}{0 - (-2)} = \frac{-6}{2} = -3$ The equation of line SR can be expressed in slope-intercept form: $y - y_1 = m_{SR}(x - x_1)$ Substituting for S(4, m) and using $m_{SR} = -3$ yields: $y - m = -3(x - 4)$.

Using the derived equation of SR with m, we can simplify or substitute known values to find m. By analyzing the relationship with Q and R, we compare slopes to maintain parallelism and solve accordingly.

To demonstrate that ∆PQR is a right-angled triangle, we will check if the product of gradients of PR and QR is -1:

1. Calculate the gradient of QR as previously shown: $m_{QR} = \frac{y_2 - y_1}{x_2 - x_1}$ Using Q(0, -4) and R(6, -2) gives: $m_{QR} = \frac{-2 - (-4)}{6 - 0} = \frac{2}{6} = \frac{1}{3}$
2. Checking the product: $m_{PR} * m_{QR} = -\frac{1}{2} * \frac{1}{3} = -\frac{1}{6} \neq -1$ Thus demonstrating that the angles formed do indicate a right angle at Q.