Deriving the distance rule by hanging a right triangle off the two points

What to assume they know

• Abscissa and ordinate, and what a coordinate pair actually records — what each coordinate measures, and separations read straight off a single axis
• Pythagoras' theorem in the form: on a right triangle, the square on the side facing the right angle equals the sum of the squares on the other two
• That a square of a number is non-negative, whatever the sign of the number
• Simplifying surds far enough to write √8 as 2√2 and √50 as 5√2
• Subtraction across zero, and squaring a negative difference
• Reading a length off a figure as a difference of two marked positions

What they should be able to do

• Set up the auxiliary construction for two given points — two perpendiculars to the horizontal axis and one horizontal segment — and name the right angle it creates
• Express each leg of that triangle as a difference of coordinates, justifying the expression from the figure rather than quoting it
• Apply Pythagoras to obtain the separation of two points in the first quadrant, and simplify the surd
• Repeat the derivation for two points in different quadrants and show that nothing in the argument changes
• Carry out the same construction with letters in place of numbers and arrive at the general rule the chapter names the distance formula
• Explain why only the non-negative square root is taken
• Derive the special case for a point measured from the origin, without treating it as a separate rule to memorise
• Explain why the rule is unchanged when the two points swap roles
• Use the rule in reverse: given a separation, solve for a missing coordinate

Where it usually goes wrong

• "You subtract the coordinates and that is the distance." It works along an axis and nowhere else. Show the two easy warm-ups from Fig. 7.2 first, then show that the same move on P(4, 6) and Q(6, 8) gives 2 and 2 and no separation at all until you square and add.
• "You have to put the smaller coordinate first." You do not, and Remark 2 is the chapter saying so. Deliberately run one of the examples in both orders on camera and land on the same surd.
• "Negative coordinates need a different formula." They need the same one. The two-quadrant example exists precisely to be worked identically to the first-quadrant one; the only skill it adds is subtracting across zero.
• "PT = 11 because you add 6 and 5." Adding happens to give the right number here only because the two points straddle the axis. Teach it as the single subtraction 6 − (−5), which also survives when both points sit on the same side.
• "The formula gives a length, so both roots are valid." A separation is a size and cannot be negative, which is the chapter's own stated reason for keeping one root. Contrast this deliberately with Exercise 7.1 question 8, where two answers are both valid — because there the unknown is a coordinate, not a length.
• "√8 is the answer." It is an answer; 2√2 is the same number written so that its size is readable. Do not attribute the habit to the chapter, which is inconsistent about it: Example 3 (p. 104) reduces all three of its radicals, Example 1 (p. 102) reduces none of its three and prints two-decimal approximations instead, and Example 2 (p. 103) leaves a reducible radical standing untouched. The same quantity, the root of fifty, appears once each way. If the explanation wants the readability lesson it has to argue for it in its own voice.
• "The right triangle is given." It is not. Nothing in the statement of the problem contains a right angle; the construction puts one there. This is the step the explanation exists to make visible.

Questions to check understanding

• Compute the separation of two given points, in one quadrant and across two
• Compute the separation of a given point from the origin
• Given two points and a stated separation, solve for a missing coordinate — and report both solutions where the quadratic yields two
• Given three points, find all three pairwise separations, as the input step to a classification question
• Justify, from the construction, why each leg equals a coordinate difference
• Explain why only the non-negative root is kept
• Word problems in the shape of the opening one: an eastward and a northward displacement, asking for the direct separation

Examples worth working on the board

Values marked verified are worked out here on data printed inside pp. 99–112; the chapter prints no answers for its exercises.

• The opening situation (§7.2, p. 100, Fig. 7.1). Town B lies 36 km to the east of town A and 15 km to its north. Fig. 7.1 draws this as a right triangle standing on the horizontal axis: the horizontal leg carries the label 36 km, the vertical leg 15 km, and the figure marks North above the vertical axis and East beside the horizontal one, and they matter, because they are what licenses treating the two displacements as perpendicular. The page names Pythagoras as the tool and then leaves the arithmetic for later. Verified: 36² + 15² = 1296 + 225 = 1521, and 1521 = 39², so the towns are 39 km apart. This value is not printed anywhere in the chapter; Exercise 7.1 question 2 (p. 105) is where the chapter comes back and asks for it.
• The easy warm-ups (§7.2, Fig. 7.2, p. 100). A(4, 0) and B(6, 0) on the horizontal axis; C(0, 3) and D(0, 8) on the vertical one. Verified: AB = 2 and CD = 5, by subtraction alone. Then the two cross-axis cases from the same figure, worked from OA = 4, OC = 3 and from OB = 6, OD = 8. Verified: AC = 5 and BD = 10. These two are already Pythagoras — one leg happens to lie along each axis, so the construction is invisible.
• The first-quadrant derivation (§7.2, Fig. 7.3, pp. 100–101). Points P(4, 6) and Q(6, 8). Construction: perpendiculars PR and QS dropped to the horizontal axis, so R is (4, 0) and S is (6, 0); then a perpendicular from P onto QS, meeting it at T. Quantities the page reads off: RS = 2, QS = 8, TS = PR = 6, hence QT = 2, and PT = RS = 2. Verified: PQ² = 2² + 2² = 8, so PQ = 2√2 ≈ 2.83. Note: this is the cleanest possible first case because both legs come out equal, which lets the audience watch the construction rather than the arithmetic.
• The two-quadrant derivation (§7.2, Fig. 7.4, p. 101). Points P(6, 4) and Q(−5, −3). Construction: QS dropped perpendicular to the horizontal axis, and a perpendicular PT from P onto QS extended. The figure marks S at (−5, 0) and labels the point where PT crosses the vertical axis as R(0, 4). So T is the point (−5, 4). The page states PT = 11 and QT = 7 and asks the reader why. Verified, and this is the answer to that question: PT runs horizontally from x = −5 across to x = 6, a span of 6 − (−5) = 11; QT runs vertically from y = −3 up to y = 4, a span of 4 − (−3) = 7. Both are differences taken across zero, which is the whole point of the example. Verified: PQ² = 121 + 49 = 170, so PQ = √170 ≈ 13.04.
• The general figure (§7.2, Fig. 7.5, p. 102). The two points are lettered now — P(x₁, y₁) below left and, further out, Q(x₂, y₂) — with PR and QS each dropped perpendicular to the horizontal axis and a further perpendicular run from P onto QS, meeting it at T. Quantities the page assembles: OR = x₁ and OS = x₂, so the gap RS comes out as x₂ − x₁, and that equals PT; SQ = y₂ and ST = PR = y₁, leaving QT as y₂ − y₁. Pythagoras on triangle PTQ then gives PQ² = (x₂ − x₁)² + (y₂ − y₁)². The rule the chapter names is the non-negative square root of that.
• Remark 1 (§7.2, p. 102). Setting one of the two points at the origin gives the separation of P(x, y) from O as the root of x² + y². Hand this over as a substitution to be performed on camera, not as a second formula.
• Remark 2 (§7.2, p. 102), which the page states and then asks the reader to justify. It writes the same rule with the two points interchanged. Verified, and this is the justification: (x₁ − x₂) is the negative of (x₂ − x₁), and squaring destroys that sign; likewise for the second coordinates. So the two expressions are equal term by term, not merely equal in value by coincidence.
• Exercise 7.1 data. (pp. 105–106), handed over as inputs. Question 1 asks for three separations: (2, 3) with (4, 1); (−5, 7) with (−1, 3); and the algebraic pair (a, b) with (−a, −b). Verified: 2√2; 4√2; and 2√(a² + b²). Question 2 asks for the separation of (0, 0) from (36, 15) and then points back at the towns. Verified: 39, so the towns are 39 km apart. Question 8 fixes P(2, −3) and Q(10, y) and states that the separation is 10 units, asking for y. Verified: 8² + (y + 3)² = 100 gives (y + 3)² = 36, so y = 3 or y = −9 — two answers, and a good place to say why a squared equation returns a pair. Question 9 fixes Q(0, 1) equally far from P(5, −3) and from R(x, 6), and asks for x and then for QR and PR. Verified: QP² = 25 + 16 = 41 and QR² = x² + 25, so x² = 16 and x = ±4; QR = √41 either way; PR = √82 when x = 4, and 9√2 when x = −4.

Figures to have open

• Fig. 7.3 redrawn as a schematic (p. 101): P(4, 6), Q(6, 8), the feet R(4, 0) and S(6, 0), and T on QS, with the right angle at T marked. The chapter's own figure; sections 4 to 6 cannot be taught without it.
• Fig. 7.4 redrawn as a schematic (p. 101): P(6, 4), Q(−5, −3), S(−5, 0), T(−5, 4), and R(0, 4) where the horizontal crosses the vertical axis. Must carry both quadrants and the axis crossing, since that is what section 8 is about.
• Fig. 7.5 redrawn as a schematic (p. 102): the same construction with letters. This is the figure the general rule is read off, and it should be visibly the same drawing as Fig. 7.3 with the numbers stripped out.
• Fig. 7.1 redrawn (p. 100): the two towns and the right triangle between them, keeping the compass labels. Standard schematic; no map is needed.
• A movement asset that morphs the numbered figure into the lettered one, so section 9 lands as a generalisation rather than a fresh start.

Where this sits in the book

• NCERT Mathematics, Class X, Chapter 7 "Coordinate Geometry", §7.2 "Distance Formula", pp. 100–102 — the town situation and Fig. 7.1, the axis warm-ups and Fig. 7.2, the first-quadrant derivation and Fig. 7.3, the two-quadrant derivation and Fig. 7.4, the general derivation and Fig. 7.5, the naming of the rule, and the two remarks that follow it.
• Exercise 7.1, questions 1, 2, 8 and 9, pp. 105–106 — the practice this topic owns. Questions 3 to 7 and 10 belong to Using distances alone to classify a triangle or a quadrilateral.
• §7.4 "Summary", p. 112, items 1 and 2 — the chapter's own restatement of the rule and of the origin case.
• Backward pointer the chapter makes itself: Pythagoras' theorem, named in §7.2, p. 100, and its converse, used from p. 103 onwards in the next topic.