Applications only. Links back to SL 2.1, equations of straight lines.
Teach the property before the recipe. Ask where a house would have to be to be equally far from both pumps, and collect answers. Most will say "in the middle". Accept it, mark the midpoint, then ask whether there is anywhere else. The second and third views answer that, and the whole line appears as the set of all such places rather than as a construction.
Then the fourth view. Do not label it as wrong first. Put it up and ask whether it could be the boundary. It passes through the midpoint, it has a sensible equation, and it looks plausible. Then read off 1.20 and 8.41. The room decides it, not you.
What they leave with is a check they own: pick a point on your answer and measure to both. That check is also the whole of 3.6.
| Question | Answer |
|---|---|
| 1. Gradient of the bisector of A(1, 2), B(7, 6) | AB has 2/3, so the bisector has −1.5. |
| 2. Its y-intercept | 4 = −1.5(4) + c, so c = 10. |
| 3. For P(−2, 5), Q(4, 1) | Midpoint (1, 3), gradient 3/2, so c = 1.5. |
| 4. Showing the student's line is wrong | B. Measure from a point on it to both sites. |
1 markThe midpoint.
1 markThe gradient of the segment.
1 markThe negative reciprocal.
1 markA correct equation.
Four marks for three one-line calculations. Insist on all three numbers being written down even by students who can do it in their heads, because a student who writes only the final equation and has the gradient wrong scores one mark out of four instead of three.
| They wrote | What happened |
|---|---|
| 1.5 | Reciprocal, no sign change. Ask them to multiply 1.5 by 2/3: they get +1, not −1. |
| −0.67 | Sign change, no reciprocal. −2/3 × 2/3 is −4/9. |
| 0.67 | The gradient of AB itself. They have not reached the perpendicular step. |
| 1.33 for the intercept | Carried the 2/3 through. The gradient error, showing up one line later. |
| −2 for the intercept | Sign slip rearranging 4 = −6 + c. |
| 4 for the intercept | Gave the midpoint's y-coordinate. Worth asking what "intercept" means. |
| 3.7 on question 3 | Used −2/3 as the bisector's gradient. The sign was already negative, so the "put a minus on it" instruction misfired. |
"It doesn't look perpendicular on my calculator." It will not, on a default window, because the x and y scales differ. Set equal scales and it snaps into place. This is worth doing once in front of them, because otherwise they distrust correct work.
"Do I have to use the midpoint? Can't I use A?" No: a line through A perpendicular to AB is a different line, and no point on it except nowhere is equidistant. Draw it and measure.
"What if AB is horizontal or vertical?" Then the bisector is the other one, and there is no gradient arithmetic. A horizontal AB gives x = the midpoint's x. Set one of these deliberately: it appears in 3.6 for every pair of sites at the same height, as it does for A(0,0) and B(8,0) there.
| Stage | What to do |
|---|---|
| Demonstrate | The equidistance check, twice, in about fifteen seconds: the distance from (0, 10) to each of A and B. Both 8.0623. Then do it for a point on the wrong line and get 1.20 and 8.41. That contrast is the demonstration, and it is faster on the machine than on paper. |
| Where they stick | Equal scales on the graph window. Also typing the distance formula without enough brackets, so the squaring happens after the subtraction only on one term. |
| The check | Ask for the product of the two gradients before anything else. If it is not −1, stop. It is a one-second test and it catches the error this sub-topic exists around. |
Degrees mode is irrelevant here, as no trigonometric function is involved, which makes this a good sub-topic to run early in the topic before the mode habit is established.
| Step | What |
|---|---|
| 1 | The two pumps. "Where is it equally far to both?" Collect answers, mark the midpoint. |
| 2 | "Anywhere else?" Views two and three. The line appears as a set of points. |
| 3 | Only now, the three steps. They are a way of finding a line they already understand. |
| 4 | View four, unlabelled. Let them judge it, then measure. |
| 5 | The negative pair, P and Q, where the perpendicular gradient is positive. |
| 6 | Segment-and-midpoint version, which is two of the three steps already done. |
| 7 | Say out loud that this is what a Voronoi edge is. It saves half of 3.6. |
Do not say "flip it and make it negative". Students apply it literally to a gradient that is already negative and produce −3/2 from −2/3. Say "the two gradients must multiply to −1", which is a test as well as an instruction and cannot misfire.
Do not let them judge perpendicularity by eye on a graph. On an unequal window it is meaningless, and a student who has once been told "it looks right" will use that instead of the arithmetic.