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

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What to assume they know

Getting the section rule from a pair of similar triangles — the section rule and the similarity argument behind it, including why the weights are crossed
Deriving the distance rule by hanging a right triangle off the two points — computing separations, needed for the checks
Solving a linear equation in one unknown that arises from clearing a fraction
Handling negative values inside a weighted sum without sign slips
Fractions with a negative numerator, and reporting an answer such as −13/3
That the diagonals of a parallelogram cut each other in half
Reading a diagram in which a physical layout is being used as a coordinate grid

What they should be able to do

Derive the mid-point expressions by putting equal parts into the section rule, rather than quoting them
Apply the section rule forwards: given two endpoints and a ratio, compute the dividing point
Find both points of trisection of a segment, and recognise that the second can be got as a mid-point instead
Recover an unknown ratio from a known dividing point, using a coordinate in which the endpoints differ and verifying with the other
Use the k : 1 form to reduce a two-unknown ratio problem to one unknown
Find the ratio in which a coordinate axis cuts a segment, by first reading off the coordinate the axis fixes
Use the mid-point rule as a test — deciding a missing vertex of a parallelogram from the fact that its diagonals bisect each other
Translate a described physical layout into coordinates and answer a distance and a mid-point question about it

Where it usually goes wrong

"The mid-point formula is a separate rule." It is one substitution away from the section rule. Deriving it takes twenty seconds and removes a whole item from the memorising list.
"Mid-point means average, so the section rule must be an average too — with the ratio on top." The section rule is an average, but a weighted one, and the weights are crossed. The mid-point case hides the crossing because the two weights are equal. Show the general case first and the special case second, in that order, or the crossing never registers.
"You need both coordinates to find an unknown ratio." You need one — any one whose two endpoint values are unequal. The second confirms. Students who try to solve both simultaneously get an identity and conclude they have gone wrong; students who reach for the coordinate the endpoints share get the same identity and are genuinely stuck, which is a different failure and worth naming separately.
"An axis cut is a different kind of problem." It is the easiest kind: the axis fixes one coordinate at zero, which is exactly the piece of information a backwards problem needs.
"A negative answer means a mistake." Example 9 lands on a negative fraction and it is correct — both endpoints sit below the horizontal axis, so the crossing must too.
"Trisection needs a new formula." It needs the same rule at 1 : 2 and then at 2 : 1, or the same rule once and a mid-point.
"AP is three sevenths of AB, so the ratio is 3 : 7." It is 3 : 4 — the ratio compares the two parts, and if three sevenths lie on one side, four sevenths lie on the other. Exercise 7.2 question 8 is built on this trap.
"A ratio of 2 : 7 means the point is near the middle." It sits two ninths of the way along, close to the first endpoint. Draw it.

Questions to check understanding

Compute the mid-point of a segment from its endpoints
Compute the dividing point from two endpoints and a stated ratio
Recover an unknown ratio from a known dividing point, and verify with the second coordinate
Find the ratio in which a named axis cuts a segment, and the crossing point
Find the points of trisection, or the three points cutting a segment into four
Given three corners of a parallelogram and the order, find the fourth
Given one end of a diameter and the centre, find the other end
Given a fraction of the whole segment, convert it to a ratio of parts before applying the rule
Layout word problems of the sports-ground type, requiring coordinates to be assigned before anything can be computed

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 mid-point collapse (§7.3, p. 107, printed as a special case). Put both parts of the ratio equal to 1 in the section rule. Verified: every weight becomes 1, the denominator becomes 2, and each coordinate of the mid-point is half the sum of the two endpoint coordinates.
Example 6 — forwards (§7.3, p. 108). Endpoints (4, −3) and (8, 5), cut at 3 : 1. Inputs only. Verified: the first coordinate is (3·8 + 1·4)/4 = 28/4 = 7 and the second is (3·5 + 1·(−3))/4 = 12/4 = 3, so the point is (7, 3). Worth saying aloud that 3 : 1 puts the point three quarters of the way along, and that 7 is indeed three quarters of the way from 4 to 8 — the arithmetic agreeing with the intuition is the reassurance students need before the backwards problems start.
Example 8 — trisection (§7.3, pp. 109–110, Fig. 7.11). Endpoints A(2, −2) and B(−7, 4); find the two points cutting the segment into three equal pieces. Fig. 7.11 is a plain horizontal strip showing A, then P, then Q, then B, with A carrying (2, −2) beneath it and B carrying (−7, 4) — the four letters and the two coordinate pairs are set inside the artwork and I read them off the printed page. Inputs only. Verified: the nearer point is at 1 : 2, giving ((1·(−7) + 2·2)/3, (1·4 + 2·(−2))/3) = (−1, 0); the further point is at 2 : 1, giving ((2·(−7) + 1·2)/3, (2·4 + 1·(−2))/3) = (−4, 2). The chapter adds a note that the second could instead be found as the mid-point of the first point and B. Verified: the mid-point of (−1, 0) and (−7, 4) is (−4, 2), which matches. Use this as the section's payoff — two routes, same answer, and the cheaper one is the mid-point rule.
Example 7 — backwards (§7.3, pp. 108–109). The point (−4, 6) is known to lie on the segment from A(−6, 10) to B(3, −8); find the ratio. Inputs only. Verified: writing the ratio as m₁ : m₂ and using the first coordinate, −4(m₁ + m₂) = 3m₁ − 6m₂ gives 7m₁ = 2m₂, so the ratio is 2 : 7. The chapter then tells the reader to confirm with the second coordinate and, on the same page, carries out that confirmation by dividing through by m₂. Verified: with the parts 2 and 7, the second coordinate comes out (2·(−8) + 7·10)/9 = 54/9 = 6, which matches. The chapter also gives the same example a second time in the k : 1 form on p. 109. Verified: −4(k + 1) = 3k − 6 gives 7k = 2, so k = 2/7 and the ratio is again 2 : 7.
Why one coordinate suffices (an added framing; the chapter demonstrates it without stating it). A point known to lie on the segment carries only one genuine degree of freedom — where along the segment it sits — so each coordinate equation says the same thing. The second is therefore a check on the data, not an extra constraint.7.
Example 9 — the cut made by an axis (§7.3, p. 110). The vertical axis cuts the segment from (5, −6) to (−1, −4); find the ratio and the crossing point. Inputs only. Verified: on that axis the first coordinate is zero, so (−k + 5)/(k + 1) = 0 gives k = 5 and the ratio 5 : 1; substituting into the second coordinate gives (5·(−4) + (−6))/6 = −26/6 = −13/3, so the crossing is at (0, −13/3). Note as a check: the minus sign on 13/3 is real. It survives on p. 110 but is dropped by text extraction, and both endpoints have negative second coordinates, so a positive answer would be impossible. Do not let a transcription lose it.
Example 10 — a mid-point used as a test (§7.3, p. 110). Four corners are named A(6, 1), B(8, 2), C(9, 4) and D(p, 3), and are stated to be, in that order, a parallelogram; find p. Inputs only. Verified: the mid-point of AC is (15/2, 5/2) and the mid-point of BD is ((8 + p)/2, 5/2), and equating the first coordinates gives p = 7. Verified observation, mine not the chapter's: the second coordinates match automatically, at 5/2 on both sides, before p is chosen at all. That is not luck — it is the problem being posed consistently, and it is worth one sentence, because a student who checks the second coordinate and finds it already satisfied should not conclude they have done something wrong.
Exercise 7.2 data (p. 111), handed over as inputs. Question 1: cut the segment from (−1, 7) to (4, −3) at 2 : 3. Verified: (1, 3). Question 2: trisect the segment from (4, −1) to (−2, −3). Verified: (2, −5/3) and (0, −7/3). Question 4: find the ratio in which (−1, 6) cuts the segment from (−3, 10) to (6, −8). Verified: 2 : 7, with the second coordinate checking out at 6. Question 5: work out how the horizontal axis cuts the segment from A(1, −5) to B(−4, 5), and give the crossing point. Verified: setting the second coordinate to zero gives k = 1, so the ratio is 1 : 1 and the crossing is the mid-point (−3/2, 0) — a satisfying collision of the topic's two halves. Question 6: (1, 2), (4, y), (x, 6) and (3, 5) are a parallelogram in that order; the unknowns x and y are wanted. Verified: equating the two diagonal mid-points gives x = 6 and y = 3. Question 7: AB is a diameter of a circle centred at (2, −3), with B at (1, 4); find A. Verified: the centre is the mid-point, so A is (3, −10). Question 8: A(−2, −2) and B(2, −4), with P on the segment placed so that AP is three sevenths of AB. Verified: that puts P at 3 : 4, giving (−2/7, −20/7). Question 9: cut the segment from A(−2, 2) to B(2, 8) into four equal parts. Verified: (−1, 7/2), then (0, 5), then (1, 13/2) — and the middle one is the plain mid-point. Question 10: find the area of the rhombus with corners (3, 0), (4, 5), (−1, 4) and (−2, −1) taken in order, the printed hint being that a rhombus's area is half the product of its diagonals. Verified: the diagonals measure √32 = 4√2 and √72 = 6√2, so the area is half of 48, namely 24 square units.
Exercise 7.2 question 3 — the sports ground (p. 111, Fig. 7.12). A rectangular school ground has corners A, B, C and D, with chalk lines ruled one metre apart and a hundred flower pots set one metre apart along the side AD. One student runs a quarter of the way along AD on the second line and plants a green flag; a second runs a fifth of the way along AD, this time on the eighth line, and plants a red one. The question asks for the gap between the flags, and then where a third student should plant a blue flag exactly half way between them. Fig. 7.12 shows the rectangle with A at the lower left, B at the lower right, D at the upper left and C at the upper right; the ruled lines run vertically and are numbered 1 to 10 along the bottom edge; the flower pots are drawn in a column up the left side; the two flags are drawn in place, one on the second line and one on the eighth. Small marks numbered 1 and 2 sit on the left edge just above A. All of that is inside the artwork and I read it off the printed page. Verified, on the reading that AD measures 100 m and the lines are numbered across from A: the green flag sits at (2, 25), the red at (8, 20), the gap between them is √(36 + 25) = √61 ≈ 7.81 m, and the blue flag goes at (5, 22.5). See the note below about the length of AD — it is a genuine ambiguity.
The boundary (A Note to the Reader, p. 112). The chapter's closing box restates the rule for a point between the two ends, and then names the other possibility — a point on the line but outside the segment — calling that dividing externally and leaving it for later study.

Figures to have open

Fig. 7.11 redrawn as a schematic (p. 109): a horizontal strip carrying A, P, Q and B in order, with A(2, −2) and B(−7, 4) labelled and the two equal-part markings shown. The chapter's own figure.
Fig. 7.12 redrawn as a schematic (p. 111): the rectangular ground lettered A, B, C, D with A lower left, ten numbered vertical chalk lines, a column of pots up the left edge, and the green and red flags in position, with room to add the blue one. The chapter's own figure and section 10 needs it; redraw rather than reproduce, and correct the vertical scale so that AD reads as a hundred metres rather than the ten the printed art suggests.
A cancellation panel showing the section rule turning into the mid-point rule, with the two identical weights struck out. Standard schematic; the chapter prints the algebra but not the picture, and section 1 is built on it.
A single segment marked at 1 : 1, at 1 : 2, at 2 : 1 and at 3 : 1, so the ratio-to-position translation can be seen at a glance. Standard schematic.
A rhombus with its two diagonals drawn and measured. Standard schematic.

Where this sits in the book

NCERT Mathematics, Class X, Chapter 7 "Coordinate Geometry", §7.3 "Section Formula" — the mid-point special case, p. 107; Examples 6 to 10, pp. 108–110; Fig. 7.11, p. 109.
Exercise 7.2, questions 1 to 10, p. 111, and Fig. 7.12.
§7.4 "Summary", p. 112, items 3 and 4 — the chapter's restatement of both the section rule and the mid-point rule.
A Note to the Reader, p. 112 — the external case, deferred.
Companion topic in this chapter: Getting the section rule from a pair of similar triangles carries the derivation this topic applies; the similarity argument is not repeated here.
Backward pointer inside this chapter: separations are computed by the rule of §7.2, p. 102, wherever a check is run.