Applications only, with no Analysis equivalent anywhere on that course.
Ask for the site of the transfer station before anything is drawn. Three depots, as far as possible from the nearest one. The room will say the middle, and several will say it confidently enough to name the centroid.
Take it seriously. Run view two and work out all three distances with them: 4.47, 4.47 and 4.00. Ask which number matters. Once somebody says "the smallest", the argument is won, because it is then obvious that the two 4.47s are wasted and you should walk away from C.
Then view three: 4.33, 4.33, 4.33. The question to ask is why it stops there. Because the moment all three are equal there is no single nearest site to walk away from. That sentence is the fourth mark on this sort of question and it is the one most often left blank.
| Question | Answer |
|---|---|
| 1. y-coordinate of the vertex | x = 4 by symmetry, and 2x + 3y = 13 gives y = 5/3 = 1.67. |
| 2. Vertex to the nearest site | √(16 + 25/9) = 13/3 = 4.33 km, to all three. |
| 3. Centroid to its nearest site | To C it is 4.00 km, which is the one that counts. |
| 4. Rainfall estimate at (1, 1) | Nearest site is A at 1.41, so 92 mm. |
| 5. Adding D(4, 1) | A. Every original cell shrinks or stays the same. |
1 markTwo correct bisector equations.
1 markSolving them for the vertex.
1 markThe distance from the vertex to a site.
1 markThe reason it is the vertex.
The reasoning mark is available on almost every version of this question and is the cheapest mark in the sub-topic. Give them the sentence to adapt: "it is equidistant from three sites, and from anywhere else you could move further from whichever site is nearest."
| They wrote | What happened |
|---|---|
| The centroid, (4, 2) | Answered the question they expected. Loses the accuracy mark and usually the reasoning mark with it. |
| 4.47 | Quoted the distance to the two furthest sites. They have the arithmetic and not the idea of "nearest". |
| 4.31 | Averaged the three distances. Worth stopping on: a mean has no meaning here at all. |
| y = 3 | Used the midpoint of A and C rather than solving two bisectors. |
| y = 2 | The centroid's y-coordinate, from averaging 0, 0 and 6. |
| 8.67 | Doubled 13/3. Usually a radius-and-diameter confusion carried in from circle work. |
| 86 mm | On question 4, C's reading. They have read the nearest LABEL rather than measured. |
| 85.3 mm | On question 4, the mean of the three. Nearest neighbour interpolation never averages, and saying why is a mark. |
| The vertex, with no distance | Stopped one line early. The question asks how far the station is from the nearest depot. |
"Why is the boundary a perpendicular bisector?" Because on it you are the same distance from both sites, so it is exactly where the nearest one changes. If 3.5 was taught from the property rather than the recipe, they will answer this themselves.
"Couldn't the best point be on an edge rather than a vertex?" Good question, and the answer is no in the bounded case the exam sets. On an edge two sites are tied and a third is nearer or further; if a third is further you can still slide along the edge away from the tied pair, and you can only stop when a third becomes tied too. The syllabus commits to it: in examinations the solution point is at an intersection of three edges.
"Do we have to draw the diagram?" Not from scratch. The guidance says coordinates of sites are given and students are not required to construct the bisectors. A rough sketch is still worth drawing, because it is the only way to see which vertex is inside the allowed region.
| Stage | What to do |
|---|---|
| Demonstrate | The simultaneous solver, with the two bisectors written as ax + by = c. Here 1x + 0y = 4 and 2x + 3y = 13, giving x = 4 and y = 1.6667. Then the distance in one entry. Point out that the vertex only needs TWO bisectors even though three meet there, and that solving the third is a free check. |
| Where they stick | Rearranging y = −2x/3 + 13/3 into 2x + 3y = 13 for the Casio's simultaneous solver, which wants that form. Also entering the fraction 13/3 as 4.33 and then losing accuracy in the next step. |
| The check | Solve the third bisector against one of the first two. If it does not give the same point, one of the three is wrong. It costs twenty seconds and it is the only independent check available. |
Keep the vertex as a fraction, 5/3 and 13/3, for as long as possible. Rounding to 1.67 early and then squaring it is the commonest source of a last-digit disagreement with the mark scheme.
| Step | What |
|---|---|
| 1 | The transfer station, cold. Collect answers. Expect the centroid. |
| 2 | View two. Work out all three distances. Ask which matters. |
| 3 | View three. Ask why it stops there. This is the reasoning mark. |
| 4 | Only now the vocabulary: site, cell, edge, vertex, each pointed at on the figure. |
| 5 | The method as three steps, and the reminder to check the region. |
| 6 | Nearest neighbour interpolation, and why it is crude. |
| 7 | View four. Adding a site, and the fact that cells can only shrink. |
Do not say "find the middle of the three sites". It is the error the sub-topic is built around, and once said it is very hard to dislodge. Say "find the point where three sites are equally near".
Do not teach the construction of bisectors with compasses here. It is not required, it eats the lesson, and it leaves students thinking the diagram is the task. The task is reading a diagram they are given and computing with it.