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

To find the gradient of NS, we identify the coordinates of N(8, 0) and S(0, -16). The gradient is calculated as:

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

The gradient (m) of line LQ can be found using the coordinates of L and Q. The change in y between L and Q is:

• From L(0, -16) to Q(4, -8): ( m_{LQ} = \frac{-8 - (-16)}{4 - 0} = \frac{8}{4} = 2. )
• Since LQ is perpendicular to SQ, we take the negative reciprocal, so: ( m_{LQ} = -\frac{1}{2}. )

Next, we use the point-slope form of a line equation: ( y - y_1 = m(x - x_1) ) Substituting in Q(4, -8): ( y - (-8) = -\frac{1}{2}(x - 4) ) This simplifies to: ( y + 8 = -\frac{1}{2}x + 2 ) ( y = -\frac{1}{2}x - 6. )

The standard equation of a circle with center at O(0,0) is given by:

$x^2 + y^2 = r^2$

To find the radius (r), we calculate the distance from O to S(0, -16): ( r = 16. )

Thus, the equation of the circle is: $x^2 + y^2 = 256.$

To calculate the coordinates of T, we use the points on line LQ. The intersection point is found by substituting the x-coordinate into the equation of LQ: Using x = 0 (since T is on the y-axis), we find: ( y = -\frac{1}{2}(0) - 6 = -6. ) Thus the coordinates of T are (0, -6).

To find the lengths LS and RS, we first calculate the distances:

• For LS, from L(0, -16) to S(0, -16): ( LS = 16 - (-8) = 8. )
• For RS, from R(6, -4) to S(0, -16): Using the distance formula: ( RS = \sqrt{(6 - 0)^2 + (-4 - (-16))^2} = \sqrt{36 + 144} = 12. ) Thus, $\frac{LS}{RS} = \frac{8}{12} = \frac{2}{3}.$

To calculate the area of trapezium PTMQ, we apply the formula for the area of a trapezium: ( Area = \frac{1}{2} (b_1 + b_2) h. ) Using base lengths between points P and T, and M and Q, where:

• Height is from PQ = 3 to 5 units. Thus, $Area = \frac{1}{2}(15)(10) = 75 \text{ square units}.$