Getting the section rule from a pair of similar triangles

What to assume they know

• Deriving the distance rule by hanging a right triangle off the two points — the perpendicular-dropping construction, and reading a leg as a difference of coordinates
• The AA similarity criterion from Chapter 6, and what "corresponding sides are in the same ratio" licenses you to write down
• Reading and manipulating a ratio written m₁ : m₂, and converting it to a single number by dividing through
• Solving a linear equation containing the unknown on both sides, including one arrived at by cross-multiplying a proportion
• That parallel lines cut by a transversal make equal corresponding angles
• The idea of a weighted average — a value pulled towards whichever input carries more weight

What they should be able to do

• Restate a positional requirement given in words as a ratio in which a point cuts a segment
• Set up the auxiliary construction of Fig. 7.9 and Fig. 7.10 and name the two triangles it produces
• Justify that the two triangles are similar, saying which two pairs of angles are used and where each comes from
• Write down the chain of equal ratios similarity gives, and identify what each term of the chain measures on the figure
• Express each of the four legs as a difference of coordinates, reading from the figure
• Derive the two coordinate expressions of the section rule by cross-multiplying and collecting the unknown
• Explain why each coordinate of the dividing point is a weighted average, and why the weights appear crossed
• Convert the two-number ratio form into the single-parameter k : 1 form and say why the two are equivalent
• State what the drawn proof assumes, and name the configuration the chapter defers to later study

Where it usually goes wrong

• "m₁ multiplies x₁." It does not, and this is the single commonest error in the whole chapter. Give students the m₁ = 0 test rather than a mnemonic: if a written formula does not collapse to A when the A-side part vanishes, it is written backwards.
• "1 : 2 means one third and two thirds of the coordinates." It means the two parts are in that ratio, so P sits one third of the way along — and the formula weights x₂ by 1 and x₁ by 2, which lands there. Show both readings and reconcile them.
• "The ratio is a fraction of the whole segment." It is a comparison of two pieces. m₁ : m₂ of 3 : 1 means three quarters of the way, not three times the segment.
• "Similar triangles give equal sides." They give equal ratios. The entire derivation exists because the two triangles are different sizes.
• "You need all three pairs of angles." The AA criterion needs two, which is what makes the derivation short. Naming the criterion by name matters — the chapter does.
• "The formula is a definition, so there is nothing to prove." The formula is a consequence of similarity. If the explanation shows only the substituting, the topic has not been taught.
• "Any point on the line AB can be reached by the rule." Only points between A and B. The chapter's own closing box says so and defers the rest.

Questions to check understanding

• Translate a worded positional condition into a ratio
• State the two triangles used in the derivation and justify their similarity
• Derive the coordinate expression for the dividing point from the similarity proportion, showing the cross-multiplication
• Given endpoints and a ratio, compute the dividing point — the direct application, treated in The midpoint as the ratio 1 : 1, and recovering an unknown ratio
• Explain why the endpoint coordinates carry the opposite parts of the ratio as weights
• Convert between the m₁ : m₂ form and the k : 1 form of the same rule
• Say what "internally" excludes, and name the case the chapter defers

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 relay tower (§7.3, p. 106, Fig. 7.9). A telephone company wants a tower at a point P lying on the segment between the two towns of §7.2, placed so that its remove from town B is twice its remove from town A. The chapter reads that requirement as P cutting the segment in the ratio 1 : 2. Frame: town A becomes the origin, and one kilometre is taken as one unit on both axes, which makes town B the point (36, 15) — carried forward from §7.2, where B lay 36 km to the east of A and 15 km to its north. Hand all of that over as inputs.
• The construction in Fig. 7.9 (§7.3, p. 106). From P and from B, drop perpendiculars to the horizontal axis, landing at D and at E; then draw PC at right angles to BE. That produces two right triangles stacked along the same slanting line: the lower one has vertices O, P and D, the upper one has vertices B, P and C. The label (x, y) is printed beside P and B(36, 15) at the far end. The pair the chapter declares similar on p. 106 is POD with BPC, by the AA criterion from Chapter 6 — not the two triangles that share the vertex O. Verified, and this part is an added filling-in: the right angle at D matches the right angle at C, and for the second pair, PC lies along the horizontal while OB cuts across both it and the axis, so the angle POD at O equals the angle BPC at P as corresponding angles. Note that BPC has no vertex at O at all, so there is nothing here that the two triangles hold in common; every equality is a matching of separate angles. PD parallel to BE gives the remaining pair, at P and at B.
• Solving the tower (§7.3, p. 106). Similarity gives two ratio statements, each equal to the ratio 1 : 2 in which P cuts the segment: one comparing the horizontal legs and one comparing the vertical legs. Written out with the coordinates of P as x and y, the horizontal statement compares x against 36 − x, and the vertical one compares y against 15 − y. Verified: 2x = 36 − x gives x = 12, and 2y = 15 − y gives y = 5, so the tower goes at (12, 5). The chapter invites the reader to confirm the position and does not perform the confirmation. Verified, and this is that confirmation: the remove from the origin is √(144 + 25) = √169 = 13, and the remove from (36, 15) is √(24² + 10²) = √676 = 26, and 13 : 26 is indeed 1 : 2. This is a strong moment — it closes the loop back to the distance topic and shows the two rules agreeing.
• The general figure (§7.3, pp. 106–107, Fig. 7.10). The two endpoints are lettered A(x₁, y₁) and, further out, B(x₂, y₂); the dividing point P(x, y) lies between them and splits the segment so that the part from P to A stands to the part from P to B as m₁ stands to m₂. Fig. 7.10 marks m₁ on the lower part and m₂ on the upper part, and labels A with (x₁, y₁), P with (x, y) and B with (x₂, y₂) — read off the printed page. Construction: from A, from P and from B drop perpendiculars to the horizontal axis, landing at R, at S and at T; then draw AQ and PC horizontally. This produces two triangles, one with A and Q as its other corners and one with B and C, and the chapter declares them similar by the same AA criterion.
• The four legs (§7.3, p. 107). Each is read as a difference of two marked positions along an axis: the horizontal leg of the lower triangle equals x − x₁; the horizontal leg of the upper triangle equals x₂ − x; the vertical leg of the lower one equals y − y₁; the vertical leg of the upper one equals y₂ − y.
• The two equations, and their solution (§7.3, p. 107). Similarity gives m₁ : m₂ equal to (x − x₁) : (x₂ − x) and also equal to (y − y₁) : (y₂ − y). Verified: cross-multiplying gives m₁x₂ − m₁x = m₂x − m₂x₁, so m₁x₂ + m₂x₁ = x(m₁ + m₂), and dividing gives x as (m₁x₂ + m₂x₁) over (m₁ + m₂). The second coordinate falls out identically with y in place of x. Verified reading: the sum of the two weights, m₁ + m₂, is the denominator, so each coordinate genuinely is a weighted average of the two endpoint coordinates — and the weight sitting on x₂ is m₁, the part nearest A, not m₂.
• A sanity check. (added here, not printed). Setting m₁ to zero puts P at A, and the formula returns m₂x₁ over m₂, which is x₁. Verified. Setting m₂ to zero puts P at B and returns x₂. So the crossing is not arbitrary — it is forced by the two limiting cases, and a student who writes the weights the other way round can catch the error in five seconds.
• The alternative derivation the chapter names but does not carry out (§7.3, p. 107). It records that the same result follows by dropping the three perpendiculars onto the vertical axis instead and repeating the argument. Worth one slide: it is the reason the two coordinates behave symmetrically.
• The k : 1 form (§7.3, p. 107). The chapter records that when the ratio is written with second part 1, the two expressions become (kx₂ + x₁) over (k + 1) and (ky₂ + y₁) over (k + 1). Verified: this is the general rule with m₁ = k and m₂ = 1, and it is legitimate because a ratio is unchanged by dividing both parts by m₂ — a move the chapter performs explicitly on p. 109.
• What is postponed (A Note to the Reader, p. 112). The closing box on the chapter's last page restates the rule, records that the ratio it refers to is the one between the two parts P makes, and then notes the other possibility: a point on the line AB but not between A and B, which is called dividing the segment externally, and which is left to later study. It is the honest boundary of the topic.

Figures to have open

• Fig. 7.9 redrawn as a schematic (p. 106): origin at A, B at (36, 15), P between them, the feet D and E on the horizontal axis, and C on BE. The chapter's own figure; sections 4 to 7 hang on it.
• Fig. 7.10 redrawn as a schematic (p. 106): A(x₁, y₁), P(x, y), B(x₂, y₂), the feet R, S and T, the horizontals AQ and PC, and the two ratio parts m₁ and m₂ marked on the segment. The chapter's own figure and the one the general rule is read off.
• A two-panel overlay showing the small triangle scaled up onto the large one, so the phrase "same shape, different size" is seen rather than asserted. Standard schematic.
• A number-line strip showing a segment cut at 1 : 2, with the two parts measured, to separate "ratio of parts" from "fraction of whole". Standard schematic; the chapter does not print it and section 2 needs it.
• No photograph or map is needed; the towns are a framing device only.

Where this sits in the book

• NCERT Mathematics, Class X, Chapter 7 "Coordinate Geometry", §7.3 "Section Formula", pp. 106–107 — the relay-tower situation, Fig. 7.9 and its two similar triangles, the general construction of Fig. 7.10, the four legs, the derivation, the naming of the rule, the note that the vertical-axis version works equally, and the k : 1 restatement.
• §7.4 "Summary", p. 112, item 3 — the chapter's own restatement of the rule.
• A Note to the Reader, p. 112 — the ratio the rule refers to, and the external case deferred to later study.
• Backward pointer the chapter makes itself: the AA similarity criterion of Chapter 6, named in §7.3, p. 106.
• Backward pointer inside this chapter: the towns and their 36-and-15 displacement, §7.2, p. 100.