The shortest route from a point to a line, read off its coefficients

0:00A point sits off to one side, and a line runs past it. How far apart are they? The question sounds harmless, but a line is not one place. Draw a few and you get a fan of segments, all different lengths. So before anything can be computed, somebody has to decide which of them the answer is. The choice made here is the smallest one. That is a definition, and a definition can be checked.

0:30In a window holding one hundred and sixty-nine points, fifty-one thousand eight hundred places on those lines were reached out to, looking for a shorter way across. Places that came out nearer than the smallest: none. Seventy-four came out exactly as near, which is the smallest being reached rather than merely approached. The shortest way across always leaves the point at a right angle to the line. Shortest is a definition. Square is a fact about it, and a fact has to be earned.

1:07Three thousand four hundred and twenty pairs of a point and a line were taken, and for each one the shortest segment was found by minimising, with no right angle assumed anywhere. Segments that turned out not to be square to their line: none. And no other point of the line offered a square segment either, so the right angle happens once and only once. From here on, distance means the shortest, and the shortest stands square.

1:40To find that shortest length you would expect to have to find where it lands. You do not. Take the point, and take the two places where the line crosses the axes. Call them the flat crossing and the upright crossing. Three points. Join them, and you have a triangle. The point is one corner, and the piece of line between the two crossings is the side opposite it. That side is a base, and the height standing over it is exactly the length we are after.

2:14So if the area of this triangle can be got twice, in two unrelated ways, the two answers can be set equal and the height falls out of the middle. The first way needs only the three corners. There is a formula that turns three pairs of coordinates straight into twice the area, and it does not care what the shape looks like. The two crossings are not free-floating. Each one is the constant divided by a leading coefficient, with a minus in front.

2:48Substitute those, and after some ordinary tidying, twice the area comes out as a single product. The constant, over the two leading coefficients multiplied together, times the size of the left-hand side evaluated at the point. That second factor is the thing everybody remembers. Put the point into the equation and see what it gives. The first factor is an awkward lump built out of all three coefficients. The second way to measure the same area is the one from primary school.

3:27Half the base, times the height. The base here is the side between the two crossings, and the height is the perpendicular from the point down onto it. The height is the number we want, so we give it a name and carry it as an unknown. Double both sides, and twice the area is the base times that unknown. The height that comes out is the height to whichever side you decided to call the base.

3:55A triangle has one area however its sides are labelled, so twice that area divided by any side returns the perpendicular height to that side, every time. Two thousand one hundred and eighty-nine of these triangles were built and checked. Ones where twice the area over the base was not the shortest way across: none, and none for either of the other two sides either. To divide by the base we need to know how long the base is.

4:30It runs from one crossing to the other, and both crossings are already written down. The distance between two points is the usual thing. Take the differences, square them, add, take the root. One crossing has nothing for its second coordinate and the other has nothing for its first, so the two differences are just the crossings themselves. The same awkward lump that turned up in the area factors straight out.

5:00What is left under the root is the sum of the squares of the two leading coefficients. So the base is that lump, multiplied by the root of the two squared coefficients added. The lump has now appeared twice. Once upstairs in twice the area, and once downstairs in the base. Divide, and it goes. Every trace of the constant, every trace of the two crossings, cancels between the top and the bottom.

5:32What survives is short enough to say in one breath. Put the point into the left-hand side, take its size, and divide by the root of the two leading coefficients squared and added. That is the whole result. It proves the lump was never part of the answer. It does not prove that some other pair of points on the line would have given the same thing, because no other pair was ever used.

6:00The reason the base could have been any side at all sits earlier, in the fact about triangles, not in this algebra. Now look back at what that derivation quietly needed. It needed a flat crossing, which exists only when the coefficient in front of the first letter is not nothing. It needed an upright crossing, which needs the second coefficient. And it needed those two crossings to be two different places, which needs the constant.

6:32If the constant is nothing, the line runs through the origin, both crossings land on the origin, the base has no length, and the division that produced the whole result is not available. Three coefficients had to be non-zero, and none of that usually gets said out loud. The derivation proves the formula for lines that cross both axes at two different places, and then the formula gets used on every line there is.

7:05Between those two sentences there is a gap, and the rest of this is spent closing it. The same window sees six thousand four hundred and sixty distinct lines. Ask the derivation to run on each in turn, and keep the ones where it refuses, because a crossing is missing or because both crossings have collapsed together. It refuses seventy-two. Now forget the derivation entirely, and describe the awkward lines out of the geometry alone.

7:40Lines every point of which stands at one height: thirteen. Lines every point of which shares one sideways position: thirteen. Lines the origin lies on: forty-eight. Those add to seventy-four, but gathered into one heap they are seventy-two, because the two axes each belong to two families at once. Lines the derivation refuses that are in none of the three families: none. Lines in a family that the derivation does not refuse: none.

8:16The same seventy-two, and neither list was built from the other. Which leaves six thousand three hundred and eighty-eight that the derivation genuinely covers. Take a level line first, one with nothing in front of the first letter. The denominator collapses to the size of the one remaining coefficient, and the whole thing reduces to the second coordinate of the point, plus the constant over that coefficient. Which is just the up-and-down gap between the point and the line, the one you could measure by eye.

8:54Three hundred and twenty-five pairings of a level line with a point were checked against that plain vertical gap. Ones where the formula gave something else: none. Turn the picture on its side and the same collapse happens for upright lines, leaving the plain sideways gap. Three hundred and twenty-five again, and again nothing disagreed. Two of the three awkward families are now covered, and not by the derivation. The third family, the lines through the origin, needs a different move.

9:32Distances do not care where the origin is. Slide the whole picture, point and line together, a few steps sideways and a few steps up, and every distance inside it is exactly what it was. But the line no longer passes through the origin, so it is now a line the derivation already covers. One thousand nine hundred and twenty such slides were performed. Slides that changed the distance: none.

10:02One hundred and forty-four of them happened to leave the line still through the origin, which is why one slide is not enough to try. So the third family follows from the case already proved, and the gap is shut. Fourteen thousand seven hundred and sixty-eight pairings of a point with a line were then scored, the formula against the smallest way across found by minimising. Disagreements: none. Fourteen thousand five hundred and sixty of those pairings were on lines the derivation covers, and two hundred and eight on lines it does not.

10:45There is a modulus in that numerator, and it is usually explained as making the answer positive. True, and it misses the point. Before the modulus closes, the left-hand side evaluated at the point has a sign, and that sign carries a fact: which side of the line the point is on. Eighty-three thousand nine hundred and ninety-three pairings had that sign read. Forty-one thousand two hundred and sixty-three came out positive.

11:18Forty-one thousand five hundred and sixty-six came out negative. One thousand one hundred and sixty-four came out as nothing at all, and those are the points standing on the line itself. Pairings where the sign failed to say which side: none. So the modulus is not tidying the answer. It is throwing a fact away, and once you know it was thrown away you can go and fetch it back.

11:47Here is the sign doing visible work. Ask for the points on the flat axis that are four units from a particular line. The modulus opens two ways, so there are two answers: at minus two, and at eight. One on each side. Ask the same question on the upright axis and you get minus eight thirds and thirty-two thirds. Anyone who treats the modulus as decoration finds one of each pair and stops.

12:18Take two lines and ask for every point that is the same distance from both. Their squared coefficients add to the same total here, thirteen and thirteen, so the condition is just that the two signed numerators have equal size. That opens two ways as well. On a grid of one thousand three hundred and sixty-nine points, seventy-three are equidistant, and seventy-three lie on one branch or the other. Equidistant points on neither branch: none. Branch points that are not equidistant: none.

12:56Each branch is itself a line, and both of them pass through the place where the original two cross. They are the two folds of the angle there, which is exactly what the modulus had been covering up. Two warnings before this gets used in anger. The first is that the coefficients must belong to the tidied equation, with everything on one side and nothing left inside a bracket. Take the point three to the right and five below the origin, and a line whose coefficients are three, minus four, and minus twenty-six.

13:36The numerator is three, the denominator is five, and the distance is three fifths. A small numerator with large coordinates: being far from the origin says nothing at all about being far from a line. Now a line handed over as one bracketed quantity equal to another. Rearranged properly, the distance from the point at minus one, one comes out as five. Reading the coefficients straight off the untidied shape describes a completely different line, one that shares not a single point of the window with the real one.

14:17Yet at that particular point it also returns five, by accident. Across the window the two agree at only thirteen places out of a hundred and sixty-nine, so do not learn the rule from whether the answer looked right. The second warning is smaller. The derivation never finds the foot of the perpendicular, so a question that asks for the foot is separate work. Cross the line with a perpendicular through the point, and that lands in exactly the same place as minimising does.

14:53Distance to a line means the shortest, which is the square one. Distance measured along some other stated direction is a different quantity altogether. Take a point and a line, and measure along a direction sloping down at forty-five degrees. The measuring line meets the given line at one place, and that squared length is eighteen. Measure squarely instead, and the squared length is two hundred and twenty-five over seventeen, which is smaller, as it has to be.

15:28Fifteen thousand two hundred and eighty directions were tried. Directions that came out shorter than measuring squarely: none. Seventy-four came out exactly equal, and every one of those was square to the line anyway. Eighty were refused outright, because a direction running alongside a line never reaches it. So: the distance is defined as the smallest, it turns out to stand square, and the coefficients hand it over without anybody ever finding where it lands.

16:03The derivation covers most lines. The formula covers all of them, but only because somebody went and checked the rest.