Method 1 (line equations). First find the main's slope from A(−2, 5) and B(6, 1): slope = (1 − 5) ÷ (6 − (−2)) = −4 ÷ 8 = −1/2. The main line is y − 1 = −1/2(x − 6), which simplifies to y = −1/2 x + 4. P lies on the main at x = 10, so y = −1/2(10) + 4 = −1, giving P = (10, −1). The branch is perpendicular to the main, so its slope is the negative reciprocal of −1/2, namely +2. The branch through P(10, −1) is y − (−1) = 2(x − 10), which gives y = 2x − 21. The reservoir sits on the y-axis, so set x = 0: y = −21.
Method 2 (vector stepping, no second equation needed). Locate P = (10, −1) as above. On the perpendicular, slope +2 means a rise of 2 for every run of 1. To go from P to the y-axis the run is 0 − 10 = −10 (ten blocks left), so the rise is 2 × (−10) = −20. The reservoir's y-coordinate is −1 + (−20) = −21. Both methods agree, so the answer is (B).
Why the wrong choices tempt: choice C (−1) is the y-coordinate of P itself, the trap of stopping at the intermediate point rather than following the branch to the reservoir. Choice D (4) comes from forgetting to turn perpendicular and reusing the main's slope, which just reproduces the main line and its y-intercept of 4. Choice E (19) keeps the wrong sign on the perpendicular slope (−2 rather than +2), the natural sign slip when taking the negative reciprocal. Choice A (−29) carries the main's constant term as −4 instead of +4, misplacing P at (10, −9) before an otherwise correct branch.