Using distances alone to classify a triangle or a quadrilateral

What to assume they know

• Deriving the distance rule by hanging a right triangle off the two points — computing the separation of two points from their coordinates
• Pythagoras' theorem and, crucially, its converse: if the squares on two sides add to the square on the third, the angle facing the third is a right angle
• The triangle inequality — any two sides together exceed the third
• Definitions carried from earlier geometry: isosceles and equilateral triangles; parallelogram, rhombus, rectangle and square, and what distinguishes each from the next
• That the points of a plane equally far from two fixed points form the perpendicular bisector of the segment joining them
• Expanding and simplifying (x − a)² algebraically, and solving a linear equation in one unknown

What they should be able to do

• Compute all three pairwise separations for a triple of points and decide from them whether the points form a triangle or lie on one line
• State the arithmetic test that separates the two cases, and say why equality is the degenerate boundary rather than a special kind of triangle
• Apply the converse of Pythagoras to three computed squares to name a right triangle, identifying which vertex carries the right angle
• Compute four sides and two diagonals of a quadrilateral and use them to name it
• Explain why equal sides alone are insufficient, and give the figure that defeats the inference
• Reproduce the chapter's shorter alternative route to the same conclusion, and say which measurements it trades against which
• Turn an equidistance condition into an equation in x and y, and recognise the resulting line as a perpendicular bisector
• Find a point on a named axis satisfying an equidistance condition, by first using the axis to eliminate one unknown

Where it usually goes wrong

• "They look collinear, so they are." Fig. 7.6 is drawn to scale and the three seats do look aligned; Exercise 7.1 question 3 supplies a triple that also looks plausible and is not. The verdict is arithmetic, never visual.
• "Sum of two sides equals the third — it is a very flat triangle." It is not a triangle. Equality is the collapse, not an extreme case of the shape.
• "Four equal sides means square." It means rhombus. The chapter's own Exercise 7.2 question 10 hands out a rhombus with four equal sides explicitly so this cannot be dodged.
• "Equal diagonals means square." On their own they mean nothing: an isosceles trapezium has equal diagonals and is no kind of rectangle. What is true needs a parallelogram already in hand — a parallelogram whose diagonals are equal is a rectangle. It is the pair of conditions that closes the case.
• "The right angle is at the vertex where the biggest number appears." It is at the vertex the longest side avoids. In Example 1 the longest is QR and the right angle sits at P.
• "Any four points make a quadrilateral." Exercise 7.1 question 6 (ii) is there to break that. Three of its four points sit on one line, so there is no four-sided figure to name.
• "Equidistant from two points" describes one point. It describes a whole line. Example 5 is what it takes to get back down to a single point: a second condition.
• "You must take square roots before comparing." You need not, and often should not — comparing squares avoids surds entirely, which is what makes the converse-of-Pythagoras route quick.

Questions to check understanding

• Given three points, decide whether they form a triangle or are collinear, and justify with the arithmetic
• Given three points forming a triangle, name it — scalene, isosceles, equilateral, right-angled — and identify the vertex carrying any right angle
• Given four points, name the quadrilateral and state which measurements force the answer
• Given four points, show that no quadrilateral is formed
• Find the relation linking x with y that makes a point equally far from two given points, and identify the resulting line geometrically
• Find the point on a named axis equally far from two given points
• Two-students-disagree questions of the Fig. 7.8 form, where the answer is a computation and a verdict, not just a number

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.

• Example 1 — triangle, and which kind (§7.2, pp. 102–103). Three points, taken as P(3, 2), Q(−2, −3) and R(2, 3). Inputs only. Verified: PQ² = 5² + 5² = 50; QR² = (−4)² + (−6)² = 52; PR² = 1² + (−1)² = 2. So PQ = √50 ≈ 7.07, QR = √52 ≈ 7.21, PR = √2 ≈ 1.41, and every pair of them sums past the third, so a triangle exists. Then the classification: 50 + 2 = 52, so PQ² + PR² = QR², and by the converse of Pythagoras the right angle sits at P — at the vertex the longest side does not touch. The chapter prints the three approximate decimals; that is a deliberate move worth copying, because it makes the inequality checkable by eye before the exact surds are compared.
• Example 3 — the classroom desks (§7.2, pp. 103–104, Fig. 7.6). Three students at desks A(3, 1), B(6, 4) and C(8, 6). Fig. 7.6 is a grid of desks drawn ten columns across and ten rows up, with the words Rows and Columns labelling the two directions and small figures marking the three occupied seats. Verified: AB = √18 = 3√2, BC = √8 = 2√2, AC = √50 = 5√2, and 3√2 + 2√2 is exactly 5√2, so the three are collinear and the students are in a line. The two shorter distances do not merely nearly reach the third, and if they fell even slightly short there would be a thin triangle instead.
• Example 2 — the square (§7.2, p. 103). Four points, taken as A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4). Inputs only. Verified: AB² = 9 + 25 = 34; BC² = 25 + 9 = 34; CD² = 9 + 25 = 34; DA² = 25 + 9 = 34 — all four sides √34. Diagonals: AC² = 4 + 64 = 68 and BD² = 64 + 4 = 68, both √68. Equal sides plus equal diagonals name it a square. Verified, and this is an added observation rather than the chapter's: 68 is exactly twice 34, which is why the second route works — 34 + 34 = 68 is the converse-of-Pythagoras statement AD² + DC² = AC². The two routes are not two coincidences; they are the same fact read twice.
• The chapter's shorter second route (§7.2, p. 103, printed as an alternative solution). Compute all four sides plus a single diagonal, then use the converse of Pythagoras on AD, DC and AC to get a right angle at D; four equal sides with one right angle is a square. Six measurements traded for five, at the cost of needing the converse.
• Fig. 7.8 and the disagreement (Exercise 7.1 question 5, p. 105). Four friends are seated at A, B, C and D on a ten-by-ten grid of desks labelled Rows and Columns; one student asserts the four make a square and the other denies it. The four seat positions are printed only inside the artwork, so they were read off a close-up of Fig. 7.8, taken wide enough to include the numbered axes and the caption, as A(3, 4), B(6, 7), C(9, 4) and D(6, 1). Hand those four pairs to the explanation as data. Verified: AB² = 9 + 9 = 18, BC² = 9 + 9 = 18, CD² = 9 + 9 = 18, DA² = 9 + 9 = 18 — all four sides √18. Diagonals: AC² = 36 + 0 = 36 and BD² = 0 + 36 = 36, both exactly 6. Sides equal and diagonals equal, so it is a square, and the student who asserted it is right. Note that the diagonals here come out as whole numbers because AC is horizontal and BD vertical — a nice contrast with Example 2, where nothing was axis-aligned.
• Exercise 7.1 questions 3, 4, 6, 7 and 10 (pp. 105–106), handed over as inputs. Question 3 asks whether (1, 5), (2, 3) and (−2, −11) are collinear. Verified: the separations are √5 ≈ 2.24, √212 ≈ 14.56 and √265 ≈ 16.28, and 2.24 + 14.56 = 16.80, which overshoots 16.28 — so they are not collinear. This is the deliberate partner to Example 3. Question 4 asks whether (5, −2), (6, 4) and (7, −2) form an isosceles triangle. Verified: two sides of √37 and one of 2, so yes. Question 6 names three quadrilaterals to classify. Verified: (i) (−1, −2), (1, 0), (−1, 2), (−3, 0) has four sides of √8 and both diagonals equal to 4 — a square. (ii) (−3, 5), (3, 1), (0, 3), (−1, −4) is a trap: the first three of those points are collinear, since stepping from (−3, 5) to (0, 3) and again to (3, 1) is the same step (3, −2) twice, so no quadrilateral is formed at all. (iii) (4, 5), (7, 6), (4, 3), (1, 2) has opposite sides √10 and √18 in matching pairs but diagonals 2 and √52 — a parallelogram and nothing more. Question 7 asks for the point on the horizontal axis equally far from (2, −5) and (−2, 9). Verified: (−7, 0). Question 10 asks for the relation making (x, y) equally far from (3, 6) and (−3, 4). Verified: 3x + y = 5.
• Example 4 — equidistance as a locus (§7.2, p. 104). A point (x, y) is to be equally far from (7, 1) and from (3, 5). Inputs only. Verified: squaring both separations and cancelling the x² and y² terms leaves x − y = 2. The chapter's own remark then reads that line back as the perpendicular bisector, which Fig. 7.7 (p. 104) draws — the figure shows A(7, 1) and B(3, 5) with the segment between them and a second line crossing it, the second line carrying its equation on a rotated label placed below the horizontal axis and close against the vertical one, on the stretch that leaves the origin heading down and to the left; no part of that label sits above the horizontal axis.
• Example 5 — one more constraint (§7.2, pp. 104–105). Find the point on the vertical axis equally far from A(6, 5) and B(−4, 3). Inputs only. Verified: a point on that axis has the shape (0, y), so one unknown is gone before any algebra; expanding gives 4y = 36 and y = 9, so the point is (0, 9), and the check AP = BP = √52 confirms it. The chapter closes the loop by noting this point is where the vertical axis meets the perpendicular bisector of AB — which is the payoff for having done Example 4 first.

Figures to have open

• Fig. 7.6 redrawn as a schematic (p. 103): a ten-by-ten desk grid with Rows and Columns labelled and three seats marked at A(3, 1), B(6, 4) and C(8, 6). The chapter's own figure; the collinearity section needs it.
• Fig. 7.8 redrawn as a schematic (p. 105): the same style of grid with four seats at A(3, 4), B(6, 7), C(9, 4) and D(6, 1). The chapter's own figure, and the positions must come from it, because the exercise prints them nowhere else.
• Fig. 7.7 redrawn as a schematic (p. 104): the two points A(7, 1) and B(3, 5), the segment between them, and the crossing line carrying its own equation.
• A rhombus-versus-square panel, both drawn with four equal sides and their diagonals shown. Standard schematic; this is the figure that carries the central argument of the topic and the chapter does not print it.
• A quadrilateral classification tree with the deciding measurement on each branch. Standard schematic.

Where this sits in the book

• NCERT Mathematics, Class X, Chapter 7 "Coordinate Geometry", §7.2 "Distance Formula", Examples 1 to 5, pp. 102–105 — the triangle test, the converse applied, the square by two routes, the desk collinearity, the equidistance relation and the point on an axis. Figures 7.6 (p. 103) and 7.7 (p. 104).
• Exercise 7.1, questions 3, 4, 5, 6, 7 and 10, pp. 105–106, and Fig. 7.8 (p. 105). Questions 1, 2, 8 and 9 belong to Deriving the distance rule by hanging a right triangle off the two points.
• Forward pointer inside the same chapter: Exercise 7.2 question 10 (p. 111) hands out a rhombus explicitly, which section 6 uses as its counter-example.
• Backward pointer the chapter makes itself: Pythagoras' theorem and its converse, the latter used from p. 103.