In the diagram, S(0; -16), L and Q(4; -8) are the vertices of \( \Delta SLQ \) having LQ perpendicular to SQ - NSC Mathematics - Question 3 - 2021 - Paper 2

To find the gradient of line NS, we use the coordinates of N(8, 0) and S(0, -16).

The formula for the gradient ( m ) is:

$m_{NS} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-16)}{8 - 0} = \frac{16}{8} = 2$

The gradient of line LQ, which is perpendicular to NS, is:\n $m_{LQ} = -\frac{1}{m_{NS}} = -\frac{1}{2}.$ To find the equation, we also use point Q(4; -8):

$y - (-8) = -\frac{1}{2}(x - 4) \Rightarrow y + 8 = -\frac{1}{2}x + 2$ So, $y = -\frac{1}{2}x - 6.$

The standard equation of a circle with center at the origin is:

$x^2 + y^2 = r^2$ Here, the radius ( r ) is the distance from O(0, 0) to S(0, -16), which is 16. Therefore, the equation becomes:

$x^2 + y^2 = 16^2 \Rightarrow x^2 + y^2 = 256.$

To find the coordinates of T, we determine the intercept of line RM with the y-axis. Using the coordinates of R and M, we calculate the y-coordinate at ( x = 0 ). Thus: [ t = (x=0, y) \text{ solved from the line equation of RM}. ]

Using the properties of triangles, we calculate the lengths LS and RS:

Let LS = QS and RS = QS - PS. Then we can set up the ratio:

$\frac{LS}{RS} = \frac{QS}{RS} = \frac{length}{length}.$

To find the area of quadrilateral PTMQ, we can break it down into triangles or trapezium. We can use the area formula:

$\text{Area} = \frac{1}{2} \times (base \times height).$ After calculating the respective coordinates and heights, we sum up the areas to achieve the final area.