The distance formula: Baudhāyana–Pythagoras rewritten in coordinates

What to assume they know

• Distance when the segment is parallel to an axis — a distance parallel to an axis as a coordinate difference
• The four quadrants, and reading a point's signs off its position — quadrants and negative coordinates
• The Baudhāyana–Pythagoras Theorem for a right triangle, from Class 8
• Squaring negative numbers, and square roots left in surd form
• Reflection of a figure in a line, as an idea of an image with corresponding points

What they 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

Where it usually goes wrong

• "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.

Questions to check understanding

• Find the distance between two given points, leaving the answer in surd form
• Find all three side lengths of a triangle from its vertices, and classify it by comparing the squares
• Decide whether three points lie on one straight line, without plotting
• Show that four given points form a square, a rectangle or a rhombus, and give the area
• Show that given points are equidistant from a stated centre, and identify the radius; then classify a further point as inside, on or outside
• Find a missing endpoint from a midpoint, and find the points of trisection of a segment
• Recover a triangle's vertices from the midpoints of its sides
• Reflect a figure in an axis and confirm which measurements are unchanged
• Applied setting: given two circular objects by centre and radius on a coordinate screen, decide whether they overlap and whether either leaves the frame

Examples worth working on the board

Values marked verified are worked out here; values marked printed are worked on the page. The chapter prints no answers to its exercises.

• Fig. 1.6, the triangle (p. 9). Axes ticked from −1 to 10 across and up to 6 in the vertical, on fine green ruling. Three points with printed coordinates: A (3, 4), D (7, 1), M (9, 6), joined into a triangle, all in the first quadrant. The chapter describes it as an acute angled triangle and asks for the lengths of AD, DM and MA.
• Think and Reflect (p. 9), two prompts that hand the student the method: how far the move from A to D covers along the x-axis and how far along the y-axis, and whether those two numbers are enough to reach AD.
• Fig. 1.7, the same triangle with the corner added (p. 10). One extra point: C (3, 1), joined to A and to D by dashed lines. Verified: C takes its x-coordinate from A and its y-coordinate from D, so AC is vertical and CD is horizontal and the angle at C is a right angle. There is a second such corner — (7, 4), taking x from D and y from A — and it gives the same two leg lengths in the other order; the chapter uses only the first. Worth one line, because it shows the construction is a choice and not a rule to memorise.
• The two legs and the hypotenuse (p. 9). Printed: CD as the x-coordinate of D less the x-coordinate of A, 7 − 3 = 4; AC as the y-coordinate of A less that of D, 4 − 1 = 3; and then AD = √(4² + 3²) = 5 units. Note: the chapter subtracts in whichever order keeps the number positive, and only later says the order does not matter.
• The other two sides (p. 10). Printed: DM = √(2² + 5²) = √29 units and MA = √(6² + 2²) = √40 units, both left in surd form. Verified: the differences come from D (7, 1) to M (9, 6), which is 2 across and 5 up, and from M (9, 6) to A (3, 4), which is 6 across and 2 down; √29 ≈ 5.39 and √40 ≈ 6.32, and √40 is also 2√10, which the chapter does not simplify.
• Checking the chapter's own description. Verified: the longest side is √40, and 40 is less than 5² + (√29)² = 25 + 29 = 54, so the largest angle is acute and the triangle really is acute angled as the page says. This is a good use of the formula: it turns a claim about a picture into arithmetic. Verified extras if wanted: perimeter = 5 + √29 + √40 ≈ 16.7 units, area 13 square units.
• Fig. 1.8, the general statement (p. 10). A right triangle drawn on the same fine ruling as the numbered figures before it, but with no axes, no arrowheads and no tick numbers — the grid is still there, the coordinate frame is what has been taken away — with three lettered vertices: A (x₁, y₁) at the top left, F (x₁, y₂) below it, D (x₂, y₂) to the right of F. The vertical leg is labelled y₂ − y₁, the horizontal leg x₂ − x₁, and the hypotenuse carries the square-root expression itself. Verified: F plays exactly the part C played in Fig. 1.7 — first coordinate from one point, second from the other.
• The sign question, and where the figure itself raises it (pp. 10–11). In Fig. 1.8, A is drawn above F, so as drawn y₂ − y₁ is a negative number while the leg it labels is a positive length. The chapter's very next sentence (p. 11) says the signs of the two differences make no difference, since what is being measured is the shift along each axis. Verified: the reason is that each difference is squared before anything else happens, so (−3)² and 3² both contribute 9. Show this with the chapter's own numbers: swapping A and D turns 4 and 3 into −4 and −3, and 16 + 9 = 25 either way.
• Fig. 1.9, the reflected triangle (p. 11). Axes ticked from −9 to 9 across. The chapter reflects the triangle in the y-axis and prints the images' coordinates: A′ (−3, 4), M′ (−9, 6), D′ (−7, 1), plus the corresponding right-angle corner C′ (−3, 1); the original C (3, 1) is lettered too. Printed: C′D′ as the x-coordinate of A′ less that of D′, −3 − (−7) = 4; A′C′ as 4 − 1 = 3; then the length √(4² + 3²) = 5 units; and D′M′ = √((−2)² + 5²) = √29, M′A′ = √((−6)² + 2²) = √40. Every length matches the original triangle, and the chapter says so.
• Think and Reflect (p. 11), two prompts: what stayed the same and what changed under the reflection, and whether the same would hold for a reflection in the x-axis instead. Verified: the three side lengths, the angles and the area are unchanged; the coordinates change sign in their first slot; the orientation of the lettering round the triangle reverses. Reflecting in the x-axis does the same thing to the second slot, for the same reason — a sign change inside a square.
• Applications the chapter sets, with the inputs intact (pp. 12–14). Each of these is a use of the formula, and the starred items are the harder ones.
  • Item 4: plot Z (5, −6), build a right triangle IZN on it, find all three side lengths — the chapter notes outright that answers will differ from person to person.
  • Item 6: are M (−3, −4), A (0, 0) and G (6, 8) on one straight line, and can it be decided without plotting? Verified: MA = 5, AG = 10, MG = 15, and 5 + 10 = 15 exactly, so the three are collinear.
  • Item 7: the same test on R (−5, −1), B (−2, −5), C (4, −12). Verified: RB = 5, BC = √85 ≈ 9.2195, RC = √202 ≈ 14.2127, and 5 + √85 ≈ 14.2195, which exceeds RC by about 0.007. So these three are not collinear — but only just, and a student rounding to one decimal place will conclude the opposite. This near-miss is the point of the pair of items, and an explanation should say so: the sum-of-distances test needs exact arithmetic, not a decimal approximation.
  • Item 9, the midpoint table (p. 13), four rows of S, M, T: (−3, 0), (0, 0), (3, 0); (2, 3), (3, 4), (4, 5); (0, 0), (0, 5), (0, −10); (−8, 7), (0, −2), (6, −3). Verified: midpoint in rows 1 and 2, not in rows 3 and 4 — row 3's midpoint is (0, −5) and row 4's is (−1, 2). The item then asks what connects M's coordinates to S's and T's.
  • Item 10: M (−7, 1) is given as the midpoint, one end is A (3, −4), and the other end B (x, y) is to be found. Verified: B = (−17, 6).
  • Item 11: the two points that cut AB into thirds, taking A (4, 7) and B (16, −2), with P the one nearer A. Verified: P = (8, 4) and Q = (12, 1).
  • Item 12: A (1, −8), B (−4, 7), C (−7, −4) all lie on a circle K centred at the origin. Verified: each is √65 from O, since 1 + 64 = 16 + 49 = 49 + 16 = 65, so the radius is √65 ≈ 8.06. Then D (−5, 6) and E (0, 9): verified 25 + 36 = 61 < 65, so D is inside; 81 > 65, so E is outside.
  • Item 13: the midpoints of the sides of a triangle are (5, 1), (6, 5) and (0, 3). Verified: the three vertices are (−1, −1), (11, 3) and (1, 7) — and which midpoint belongs to which side does not change that set.
  • Item 15: a screen 800 pixels wide and 600 high, origin at the bottom-left; one circular icon of radius 80 centred at A (100, 150), another of radius 100 centred at B (250, 230). Verified: the first spans x from 20 to 180 and y from 70 to 230, the second x from 150 to 350 and y from 130 to 330, so both sit entirely on the screen; AB = √(150² + 80²) = √28900 = 170 exactly, which is less than 80 + 100 and more than 100 − 80, so the two circles cross at two points.
  • Item 16: A (2, 1), B (−1, 2), C (−2, −1), D (1, −2). Verified: all four sides are √10, both diagonals are √20, so ABCD is a square of area 10 square units. Be careful how the diagonals are used here: a square is a rhombus, so equal diagonals cannot rule the rhombus family out. What they rule out is a rhombus that is not a square — they force the angles to be right.

Figures to have open

• Fig. 1.6 redrawn (p. 9): the triangle A (3, 4), D (7, 1), M (9, 6) on ticked axes. The chapter's own figure.
• Fig. 1.7 redrawn (p. 10): the same triangle with C (3, 1) and the two dashed legs. The chapter's own figure, and the argument of sections 3–5 is this picture.
• Fig. 1.8 redrawn (p. 10): the general right triangle with A (x₁, y₁), F (x₁, y₂), D (x₂, y₂), the legs labelled by their differences and the hypotenuse by the root. The chapter's own figure; keep the leg label as y₂ − y₁, because section 8 is about exactly that mismatch between a signed label and a positive length.
• Fig. 1.9 redrawn (p. 11): both triangles either side of the y-axis with all primed coordinates. The chapter's own figure.
• A circle centred at the origin with radius √65 and the five points of item 12 plotted, for section 11. Standard schematic; the chapter prints no figure for it.
• A pixel-screen rectangle 800 by 600 with the two icons of item 15 drawn to scale. Standard schematic; nothing printed.

Where this sits in the book

• NCERT Ganita Manjari Class 9 (Part I), printed Chapter 1, §1.4 "Distance Between Two Points in the 2-D Plane", pp. 8–11, with Fig. 1.6 (p. 9), Fig. 1.7 and Fig. 1.8 (p. 10) and Fig. 1.9 (p. 11), and the two Think and Reflect boxes on pp. 9 and 11
• End-of-Chapter Exercises items 4, 6, 7, 9, 10, 11, 12, 13, 15 and 16, pp. 12–14, including the midpoint table on p. 13
• The Chapter Summary restates the formula, attributed to the Baudhāyana–Pythagoras Theorem, at p. 15
• The theorem is named in the chapter's opening history at §1.1, p. 1, and is taken as known from Class 8