The midpoint as the ratio 1 : 1, and recovering an unknown ratio

0:00The midpoint rule is not a second formula. It is the section rule with one substitution made. Here is the rule. The first coordinate is m-one times x-two, plus m-two times x-one, all over m-one plus m-two. Now make the two parts equal. Call them both m, and substitute. The numerator becomes m x-two plus m x-one, and the denominator becomes two m. Take the m out of the top, and it cancels, leaving x-one plus x-two over two.

0:37The second coordinate does the same thing, giving y-one plus y-two over two. That is the midpoint rule, arrived at by one line of cancelling rather than from memory. Look at what cancelled. In the general rule the two weights are crossed: the coordinate of the far endpoint carries the part nearest the other end. When the two parts are equal, that crossing is still there. It has not gone away. It has become invisible, because swapping two identical weights changes nothing.

1:14That is why the midpoint is a bad place to learn the section rule from: every mistake about which weight goes where gives the right answer here. I measured that over ten thousand four hundred and forty cases: eight hundred and seventy pairs of endpoints and twelve ratios. The rule lands on the plain average on the one thousand seven hundred and forty where the two parts are equal, and on none of the other eight thousand seven hundred.

1:45And the point it names is the midpoint by definition, not by formula: equally far from both ends, on the line, between them. A point one unit off failed that on every pair. Now use the rule forwards. The endpoints are four, minus three, and eight, five, and the ratio is three to one. First coordinate. Three times eight, plus one times four, is twenty-eight, over four. That is seven.

2:17Second coordinate. Three times five, plus one times minus three, is twelve, over four. That is three. So the point is seven, three. Now the check. Three parts then one is four parts, so the point should sit three quarters of the way along. From four to eight is a run of four, and three quarters of four is three. The arithmetic and the picture agree, and that check is the one thing that catches a sign error when we run the rule backwards.

2:52Next, cutting a segment into three equal pieces: the two points of trisection. The endpoints are two, minus two, and minus seven, four. The nearer point has one piece behind it and two ahead, so it sits at one to two. One times minus seven, plus two times two, is minus three, over three, which is minus one. The second coordinate comes out at nothing. So the first point is minus one, nought.

3:24The further point has two pieces behind and one ahead, so it sits at two to one, which gives minus four, two. Now the shortcut. Once you have the first point, the second is simply the midpoint of it and the far end. Minus one plus minus seven, halved, is minus four. Nought plus four, halved, is two. The same answer, from a rule that needs no ratio at all.

3:53Now turn the question round. The point minus four, six lies on the segment from minus six, ten to three, minus eight. In what ratio does it cut it? Write the ratio as m-one to m-two and put the numbers into the first coordinate. Minus four equals three m-one minus six m-two, all over m-one plus m-two. Clear the fraction. Minus four m-one minus four m-two equals three m-one minus six m-two.

4:27Gather, and seven m-one equals two m-two, so the ratio is two to seven. Two parts near, seven far, nine in all, so the point sits two ninths of the way along. Nowhere near the middle: a ratio of two small numbers always looks more central than it is. Now confirm with the second coordinate, and understand why that is a confirmation rather than a second equation. Six equals minus eight m-one plus ten m-two, over m-one plus m-two.

5:04Clear and gather, and fourteen m-one equals four m-two, which is seven m-one equals two m-two. The same equation. Not a similar one. The same one. Here is the reason. A point on the segment has one thing left to decide: how far along it sits. One number, not two. So each coordinate asks the same question. Over all ten thousand four hundred and forty cases, whenever both coordinates could answer, they agreed, and agreed with the ratio the point was built from.

5:41But a check that always agrees is worth nothing, so I shifted each answer one unit off the segment. On all seven thousand two hundred cases where both could still answer, they disagreed. So the second coordinate is not testing your algebra. It is testing whether the point you were handed lies on the segment at all. There is one thing to be careful about that hardly ever comes up, because most segments you meet slant to both axes.

6:15Suppose the segment stands exactly upright. Both endpoints have the same first coordinate, and so does every point between them. Put that into the first-coordinate equation and the unknown disappears, leaving a true statement with no ratio in it. Nothing has gone wrong. That coordinate simply had nothing to say, and the second one still answers perfectly. A segment lying exactly flat does the mirror image. Of the eight hundred and seventy pairs, a hundred and twenty stand upright and a hundred and fifty lie flat, leaving six hundred oblique to both.

6:59On the upright ones only the second coordinate can answer, on the flat ones only the first, and on the oblique six hundred both can and both agree. So the rule is: pick a coordinate in which the two endpoints actually differ. A student who reaches for the coordinate the two ends share has not made an algebra mistake. They have picked an equation with no unknown in it, which needs a different kind of help.

7:30A ratio does not change if you divide both parts by the same number, so divide both by m-two. The ratio becomes m-one over m-two, to one. Call that single number k, and the ratio is k to one. The rule becomes k x-two plus x-one, over k plus one. One unknown instead of two. The same problem: minus four equals three k minus six, over k plus one. That gives two equals seven k, so k is two sevenths.

8:06So k to one is the ratio two to seven. The same answer, with less writing. I checked the two forms against each other on all ten thousand four hundred and forty cases. They give the same point every time. Now the easiest kind of backwards problem. In what ratio does the upright axis cut the segment from five, minus six to minus one, minus four, and where does it cross?

8:37The thing to notice is that you have already been given a coordinate: every point on the upright axis has first coordinate nought. So put nought in. In the one-letter form, minus k plus five, over k plus one, is nothing. A fraction is nothing when its top is, so k is five and the ratio is five to one. Now the crossing, from the second coordinate. Five times minus four, plus minus six, is minus twenty-six, over six. That is minus thirteen thirds.

9:15And that minus sign is real, not a slip. Both endpoints sit below the flat axis, so anything between them must too. A positive answer would have been impossible. I ran the same move on every pair in the population and it lands on the axis every time. Taking the ratio from the other coordinate instead misses the axis on hundreds of them. A point P sits on the segment from A to B, placed so that A-P is three sevenths of A-B. It is tempting to write the ratio as three to seven, and it is wrong.

9:57Three sevenths of the way along leaves four sevenths to go, so the ratio is three to four. Three to seven would mean ten parts in all, and a different point. With A at minus two, minus two and B at two, minus four, the honest reading gives minus two sevenths, minus twenty over seven. The careless one gives minus four fifths, minus thirteen fifths. I ran both readings over every pair of endpoints and eight different fractions. They never once landed in the same place.

10:34A fraction measures a piece against the whole. A ratio measures a piece against the other piece. Convert before you compute. Now use the midpoint rule as a test. Four corners in order: six, one; eight, two; nine, four; and p, three, stated to be a parallelogram. Find p. The fact that does the work is that the diagonals of a parallelogram cut each other in half. They join the first corner to the third and the second to the fourth.

11:09Midpoint of the first is fifteen halves, five halves. Midpoint of the second is eight plus p over two, and five halves again. Set the first coordinates equal. Eight plus p is fifteen, so p is seven. Now something worth pointing out. Look at the second coordinates: five halves on both sides, and p never appeared in either. They matched before we chose anything. That is not luck and not a mistake. It is the question being posed consistently, and it happened on all one thousand one hundred and eighty-two triples I tried.

11:51And the corner those bisecting diagonals hand you really does close a parallelogram, checked by comparing displacements rather than midpoints. Reading the wrong pair as the diagonals closes it on none. A rectangular school ground has chalk lines ruled one metre apart and a hundred flower pots set one metre apart along one side. One student walks a quarter of the way along that side and plants a green flag on the second chalk line. Another walks a fifth of the way and plants a red one on the eighth.

12:31Nothing can be computed until the layout becomes coordinates, and that translation is the whole difficulty. Take the corner where the pots start as the origin, number the chalk lines across, and measure the pots up. I will read that side as a hundred metres. A quarter of it is twenty-five and a fifth is twenty, so the green flag is at two, twenty-five and the red one at eight, twenty.

13:01The gap is six across and five up, so sixty-one, and the distance is the root of sixty-one, a little under seven point nine metres. Half way between them is the midpoint, five, twenty-two and a half, and that is where the blue flag goes. One honest note. A hundred pots a metre apart span ninety-nine metres, not a hundred, and on that reading the flags sit lower and the gap is a different number.

13:32I have taken the hundred, because the quarter and the fifth are plainly meant to come out whole. Say which reading you are using. Four corners in order: three, nought; four, five; minus one, four; and minus two, minus one. They form a rhombus. The shortcut for its area is half the product of its two diagonals. First diagonal, from three, nought to minus one, four: four across and four up, so its square is thirty-two. Second diagonal, from four, five to minus two, minus one: six and six, so its square is seventy-two.

14:14Neither has a whole square root, and you need neither on its own. Multiply the two squares first: thirty-two times seventy-two is two thousand three hundred and four, whose root is forty-eight exactly. Half of that is twenty-four square units. Now, is the shortcut about rhombi, or about something else? I checked it against an area computed a completely different way, over three families. Sixty-four rhombi: it agreed on all of them. Twenty-seven kites, which have two pairs of equal sides and no more: it agreed on those too. Twenty-seven figures whose diagonals do not cross at right angles: it was wrong on every one.

15:03So the shortcut is not really about rhombi. It is about diagonals meeting at right angles, and a rhombus is simply one figure where they do. Two things to take away, and then the boundary. First, the midpoint rule is the section rule with the two parts made equal, and that cancellation is why the crossed weights are invisible there. Second, running the rule backwards needs one coordinate, not two: any coordinate in which the two endpoints differ. The other confirms the data.

15:41Now the boundary. Everything here has assumed the dividing point lies between the two ends, and that is not a small assumption. Feed the rule a ratio with a negative part and it still hands you a point on the line through the two ends. On all eight hundred and seventy pairs: on the line every time, between the ends on none. That situation has a name: the point divides the segment externally, and it is left for later study.

16:13So the rule is complete for the case it covers and silent about the one it does not, and knowing which is which is most of what it means to know a rule.