1. Given that points P, Q, and S lie on a Cartesian plane with PQ parallel to the x-axis and PQ equals 20 times the length of segment S. 2. Point S is at (-6,0), Q is at (k,8), and line PQ is parallel to the x-axis so P and Q have the same y-coordinate. Therefore, P is at (x_P, 8). 3. The distance PQ along the x-axis is $|k - x_P|$. 4. Since PQ is parallel to x-axis, P and Q share the same y-value (8), so PQ is the horizontal distance. 5. The length of segment S is the distance from S to origin O: $OS = \sqrt{(-6-0)^2 + (0-0)^2} = 6$. 6. Given PQ = 20S, we have $|k - x_P| = 20 \times 6 = 120$. 7. Since P lies on the line with y=8 and on the vertical line passing through S (x = -6), P must be at (-6,8). 8. Thus, $|k - (-6)| = 120$ which simplifies to $|k + 6| = 120$. 9. Solve for k: $k + 6 = 120$ or $k + 6 = -120$. 10. $k = 114$ or $k = -126$. 11. Considering location of Q above x-axis and plausible context, $k = 114$. (b) To find the equation of line PS: 1. Coordinates: P = (-6,8), S = (-6,0). 2. Find the slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0-8}{-6+6} = \frac{-8}{0}$ which is undefined. 3. Line PS is vertical at x = -6. Therefore, the equation of line PS is: $$ x = -6 $$