Chapter 7.1 — Exercise 7.1 — Distance Formula

Building the Formula From a Right Triangle

Take two points A(x₁,y₁) and B(x₂,y₂). Drop a perpendicular from each down to the x-axis, and let C be the point directly below A and level with B. That construction quietly builds a right triangle: AC runs horizontally with length x₂−x₁, BC runs vertically with length y₂−y₁, and the angle at C is exactly 90°. AB is the hypotenuse of that triangle, so Pythagoras' theorem hands over the distance directly.

When one of the two points is the origin O(0,0), the formula collapses to something even shorter: the distance from the origin to any point (x,y) is simply √(x²+y²), since x₁ and y₁ both drop out as zero.

Four Straight Distances

The last row is worth pausing on, since it's the only one built from letters instead of numbers: the formula doesn't care whether x₁, y₁, x₂, y₂ are specific digits or unknowns, it just needs them substituted in the same order every time. The distance between (0,0) and (36,15) uses the origin shortcut directly: √(36²+15²) = √(1296+225) = √1521 = 39 units — a clean whole-number answer that's rare enough in this exercise to be worth noticing when it happens.

When Three Points Refuse to Line Up

Checking whether three points are collinear doesn't need a ruler laid across the page — it needs exactly one geometric fact: three points are collinear only when the distance between the two outer points equals the sum of the distances to whichever point sits between them. For (1,5), (2,3), (−2,−1): AB=√5, BC=√32=4√2, AC=√45=3√5. Since AB+BC (≈7.89) doesn't equal AC (≈6.71), the three points are not collinear — they form a genuine triangle instead of a straight line.

Naming Shapes by Their Sides Alone

Once every side and diagonal of a quadrilateral can be measured from coordinates, naming the shape becomes a matter of comparing lengths rather than looking at a picture:

A rectangle and a rhombus both start from "opposite sides equal" or "all sides equal," but only checking the diagonals too tells them apart from a plain parallelogram — equal diagonals alongside equal opposite sides is what actually locks in a rectangle, and the same equal-diagonal check upgrades a rhombus into a square.

Four Shapes, Confirmed by Distance

• Isosceles triangle: (5,−2), (6,4), (7,−2) — AB=√37, BC=√37, AC=2. AB=BC, so these three points genuinely form an isosceles triangle.
• Classroom seating (Jarina and Phani): four friends sit at A(3,4), B(6,7), C(9,4), D(6,1). Jarina claims ABCD is a square; Phani disagrees. Checking: AB=BC=CD=DA=√18, and both diagonals AC=BD=6. All sides equal and both diagonals equal — Jarina is correct, ABCD is a square, settled by arithmetic rather than argument.
• Equilateral triangle: A(a,0), B(−a,0), C(0,a√3) — every side works out to 2a regardless of what value a actually takes, confirming an equilateral triangle for any nonzero a.
• Parallelogram: (−7,−3), (5,10), (15,8), (3,−5) taken in order — AB=CD=√313 and BC=AD=√104, opposite sides equal in pairs, confirming a parallelogram.

A Rhombus With a Measurable Area

Points (−4,−7), (−1,2), (8,5), (5,−4), taken in order, all have equal side length √90 — a rhombus. But a rhombus's area isn't found from its sides at all; it comes from its diagonals instead, since a rhombus's diagonals always cross at right angles and split it into four equal right triangles.

Here AC=√288=12√2 and BD=√72=6√2, so the area is ½ × 12√2 × 6√2 = ½ × 144 = 72 square units.

Three Quadrilaterals, Named by Their Diagonals

The second and third rows look almost identical in shape (equal opposite sides, unequal diagonals) — both stop at parallelogram and go no further, since it's precisely the unequal diagonals that rule out anything more specific like a rectangle.

Finding an Unknown Point or Coordinate

Several problems in this exercise flip the usual question: instead of measuring a distance between two given points, they hand over a distance and ask what point or coordinate makes it true.

• A point on the x-axis, equidistant from (2,−5) and (−2,9): any point on the x-axis has the form (x,0). Setting its distance to both given points equal: (x−2)²+25 = (x+2)²+81, which simplifies to −8x=56, so x=−7. The point is (−7, 0).
• Find x, given the distance between (x,7) and (1,15) is 10: (x−1)²+64=100 leads to (x−1)²=36, so x−1=±6. x = 7 or x = −5 — both values are genuinely valid, since a distance of 10 can be reached moving in either direction along x.
• Find y, given the distance between (2,−3) and (10,y) is 10: 64+(y+3)²=100 leads to (y+3)²=36. y = 3 or y = −9, for the same reason as above.
• The radius of a circle centred at (3,2) and passing through (−5,6): the radius is just the distance from centre to that point on the circle. r=√[(−5−3)²+(6−2)²]=√80 = 4√5 units.

A Triangle That Refuses to Exist

Not every set of three points can form a triangle — the distance formula catches this before any attempt at drawing one. For (1,5), (5,8), (13,14): AB=5, BC=10, AC=15. Since AB+BC=15 exactly equals AC, the three points sit on a single straight line rather than at the corners of a triangle. No triangle can be drawn with these three vertices — they're collinear, the same test used earlier in this exercise, just applied here to explain an impossibility rather than confirm a shape.

An Equation Instead of a Number

One further variation asks for a relationship rather than a single value: find the condition linking x and y such that (x,y) is always equidistant from (−2,8) and (−3,−5). Setting the two distances equal and squaring both sides:

Every point satisfying this single linear equation — not just one specific (x,y) pair — sits exactly halfway between the two given points, tracing out the perpendicular bisector of the segment joining them.

From Distance to Division

Every problem in this exercise measured the straight-line gap between two points. Exercise 7.2 asks a related but different question: not how far apart two points are, but exactly where a third point sits if it divides the segment between them in some given ratio — a formula built directly on the same coordinate-pair thinking used throughout this exercise.