PrepShorts · Study sheet · Class 9 Mathematics · Chapter 1, Orienting Yourself: The Use of Coordinates
Chapter 1 · Orienting Yourself: The Use of Coordinates
The distance formula: Baudhāyana–Pythagoras rewritten in coordinates
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The distance formula is not a new fact. It is Baudhāyana–Pythagoras with the two legs written as coordinate differences.
The idea
Two points anywhere in the plane are already two corners of a right triangle — you get the third for free by taking one point's first coordinate and the other point's second — and the two legs of that triangle are axis-parallel, so their lengths are nothing but coordinate differences. That is the whole content of the distance formula: Baudhāyana–Pythagoras with the legs written in coordinates, and no new measurement anywhere. And because the legs enter squared, the sign of each difference is discarded, so it cannot matter which of the two points the subtraction starts from — which is why one formula serves all four quadrants and why the chapter's reflected triangle comes out with exactly the side lengths it started with. Note the scope: it is the sign of a difference that squaring erases, not the sign of a coordinate. Move a point from (1, 0) to (−1, 0) and its distance from (2, 0) changes from 1 to 3.
What you should be able to do
- Given two points, construct the third vertex that makes a right triangle with legs parallel to the axes, and give its coordinates
- Compute each leg as a coordinate difference and obtain the slanted length by the Baudhāyana–Pythagoras Theorem
- Reproduce the chapter's three side lengths for its triangle, leaving irrational answers as square roots
- Write the general formula for the distance between two named points, and say which part of the figure each difference corresponds to
- Explain why the sign of each difference cannot affect the result, and why that makes a single formula sufficient for all four quadrants
- Compute distances when some or all coordinates are negative, and confirm that reflecting a figure in an axis leaves every side length unchanged
- Use distances to decide whether three points lie on one straight line, whether four points form a square, and whether a point is inside, on or outside a circle
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| Baudhāyana–Pythagoras Theorem | in a right triangle, the square on the longest side equals the sum of the squares on the other two | printed in this chapter, §1.1, p. 1, and applied at §1.4, p. 9 |
| acute angled triangle | a triangle all of whose angles are less than a right angle | printed in this chapter, §1.4, p. 9 |
| right-angled triangle | a triangle with one angle of ninety degrees | printed in this chapter, End-of-Chapter Exercises item 4, p. 12 |
| reflection | the transformation that takes a figure to its mirror image in a line | printed in this chapter, §1.4, p. 11 |
| image | the point a given point is carried to by the reflection, written with a prime | printed in this chapter, §1.4, p. 11 |
| midpoint | the point of a segment equidistant from both ends | printed in this chapter, End-of-Chapter Exercises item 9, p. 12 |
| points of trisection | the two points cutting a segment into three equal parts | printed in this chapter, End-of-Chapter Exercises item 11, p. 13 |
| distance formula | the expression giving the distance between two points from their coordinates | an added term; the chapter states the expression, at p. 10 and again in the summary at p. 15, without giving it a name |
| collinear | lying on one straight line | not printed in this chapter — the explanation's word for what items 6 and 7 call being on the same straight line; the word is printed elsewhere in this volume |
| leg | one of the two shorter sides of the right triangle, each parallel to an axis | an added word in this chapter, which names those two sides only by their endpoints |
Where people slip up
- "The distance formula is a new fact to learn." It is the Class 8 theorem with the legs written as coordinate differences. If a student can build the third corner, they can rebuild the formula from scratch, which is the only reliable way to remember it.
- "You subtract in the order the points are named." You may subtract either way. The chapter itself takes the x-difference one way round and the y-difference the other, precisely because both give the same square.
- "With negative coordinates you need a different version of the formula." The reflected triangle in Fig. 1.9 is the chapter's demonstration that you do not: −3 − (−7) is 4 exactly as 7 − 3 is 4, and every length comes out unchanged.
- "Squaring a negative gives a negative." This is the specific error the whole argument rests on. (−2)² = 4 and (−6)² = 36, as the chapter's own reflected calculations print.
- "√29 should be turned into 5.39." The chapter leaves it as √29, and for a length that is the exact answer; the decimal is an approximation. Item 7 shows what the approximation costs — rounding turns a non-collinear triple into a collinear-looking one.
- "If the three distances nearly add up, the points are on a line." Nearly is not a mathematical verdict. Item 7's triple misses by about seven thousandths of a unit, and it is a genuine miss.
- "Any four points with equal sides make a square." Equal sides make a rhombus. Item 16 needs the diagonals too, and they are equal there.
- "The formula tells you the direction as well as the distance." It returns a single non-negative number. Everything about which way round the two points lie has been squared away.
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Worked answers to this chapter’s exercises · this video explains End-of-Chapter Exercises Q3, End-of-Chapter Exercises Q4, End-of-Chapter Exercises Q6, End-of-Chapter Exercises Q7, End-of-Chapter Exercises Q8, End-of-Chapter Exercises Q12, End-of-Chapter Exercises Q15, End-of-Chapter Exercises Q16
Transcript1,297 words
Two points that share a coordinate are the easy case, and we have done it. Here are two that share nothing. The segment between them slants, and there is no axis lying along it to subtract on. There is no third rule coming for this, and you would not want one. A separate rule for slanted segments would only be a separate thing to forget. What there is, is a right triangle you already own. And you can build it in a single move, out of numbers you have already been given.
Here is a triangle, handed over the way this whole subject hands things over: as three pairs of numbers. A at three, four. D at seven, one. M at nine, six. Three sides, and every single one of them slants. Not one lies along an axis, so not one of them can be read off by subtracting. And yet all three lengths are already fixed. Nothing here is waiting to be measured.
Take the first coordinate of A, and the second coordinate of D. Three, and one. Put a point there and call it C. Now look at what that does. C shares its first number with A, so the segment up to A is vertical. C shares its second number with D, so the segment across to D is horizontal. Vertical meets horizontal at a right angle. You did not find that corner anywhere. You manufactured it.
And there is a second one, at seven, four, taking the first number from D and the second from A. Either will do. This is a choice, not a rule to memorise. The corner costs nothing, because both of the numbers it is made of were already sitting in front of you. The two new sides are now the easy kind, which is the entire point of building them. The bottom one runs from three, one across to seven, one. Same second number, so subtract the firsts. Seven minus three is four.
The upright one runs from three, one up to three, four. Same first number, so subtract the seconds. Four minus one is three. Three and four, and no ruler has been anywhere near this picture. Both of them came straight out of the coordinates. And now a theorem you have had for a year finishes the job. In a right angled triangle, the square on the longest side is the sum of the squares on the other two.
Four squared is sixteen. Three squared is nine. Sixteen and nine make twenty five, and the square root of twenty five is five. The slanted side is five. That is the distance formula. Not a new fact, but that old theorem with its two legs written as coordinate differences. Which means anybody who can build the corner can rebuild the formula from nothing, and that is the only reliable way to remember anything.
Go round the triangle twice more. Same three moves each time. From D to M is two across and five up. Four and twenty five is twenty nine. From M back to A is six across and two down. Thirty six and four is forty. Twenty nine and forty are not perfect squares, so those two sides are the square root of twenty nine and the square root of forty.
Leave them like that. A square root is the exact length. A decimal is a rounding of it, and shortly you will see what the rounding costs. Here is something worth doing with three lengths you have just built. This looks like a triangle with no blunt corner in it. Looks like. The longest side is root forty, so its square is forty. The other two squares are twenty five and twenty nine, and those add to fifty four.
Forty is less than fifty four, which makes the largest angle smaller than a right angle. So every angle in it is. A claim about a picture has just turned into arithmetic. That is the formula doing work, rather than restating itself. And notice you never needed the angles themselves. Three squared lengths settled the whole question. Now do the whole thing once with letters, and you never have to do it again.
One point at x one, y one. The other at x two, y two. Build the corner exactly as before. First coordinate from one point, second coordinate from the other. The horizontal leg is x two minus x one. The upright leg is y two minus y one. Square both, add them, take the root. There is nothing in that formula that was not already in the picture. The letters are not abstraction for its own sake. They are the same two subtractions, done once instead of every single time.
Now look hard at that upright leg, because as drawn there is something wrong with its label. The point on top is the one with the larger second number, so y two minus y one comes out negative, and it is labelling a side whose length is positive. That mismatch does not matter, and it is worth knowing exactly why rather than being told to ignore it. Each difference gets squared before anything else happens to it. Minus three squared is nine. Three squared is nine.
Swap the two points over and the differences become minus four and minus three. Sixteen and nine. Still twenty five. So the order you subtract in cannot reach the answer at all. One formula covers all four quadrants. But be careful what that does and does not say. It is the sign of a difference that gets squared away. Not the sign of a coordinate. A point at one, zero sits one unit from two, zero. Move it to minus one, zero and it is three units away. Negating a coordinate changed the distance.
What does survive is a mirror. Reflect the whole triangle across the upright axis. A goes to minus three, four. D to minus seven, one. M to minus nine, six. Minus three minus minus seven is four. The other leg is still three. Five again, and root twenty nine, and root forty. Every length unchanged. Mirroring across the flat axis does the same thing to the second slot, for exactly the same reason.
One more thing the formula settles, with one trap sitting inside it. Three points lie on one straight line exactly when the two shorter distances add up to the long one. Take a point at minus three, minus four, the origin, and a point at six, eight. The distances are five, then ten, and fifteen end to end. Five and ten make fifteen, so they are on a line, with no plotting at all.
Now three others. Minus five, minus one. Minus two, minus five. Four, minus twelve. The distances are five, root eighty five, and root two hundred and two. Round those to one decimal place and you get fourteen point two, twice. It looks like a line. It is not one. Five plus root eighty five is larger than root two hundred and two, by less than a hundredth of a unit, but larger. Nearly is not a verdict.
One construction, one theorem, one formula. Whether three points are on a line, or very nearly on a line. Whether four points make a square, which needs the diagonals checked as well as the sides. Whether points are all the same distance from a centre, which is what a circle is. Every one of those is this. Any two points are already two corners of a right triangle. The third corner is free.
And the two legs were never measured. They were subtracted. None of this was a new idea. It is one old theorem, read in coordinates.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Distance when the segment is parallel to an axisClass 9 · Ch 1, Orienting Yourself: The Use of Coordinates
- The four quadrants, and reading a point's signs off its positionClass 9 · Ch 1, Orienting Yourself: The Use of Coordinates
Either side of this one
- Turning a situation into an expression: terms, variables, coefficientsClass 9 · Ch 2, Introduction to Linear Polynomials