Jugami Geometry

Jugami › Geometry › Coordinate Geometry

Coordinate Geometry

Put a grid under a figure and every point becomes a pair of numbers. Then lengths, midpoints, right angles and circles become arithmetic.

This unit covers the distance formula (the Pythagorean theorem on a grid), the midpoint formula, slope with the tests for parallel and perpendicular lines, the equation of a circle, and coordinate proofs.

Points A at 1, 1 and B at 5, 4 on a coordinate grid. The segment AB is the hypotenuse of a right triangle with a horizontal leg of 4 and a vertical leg of 3, so AB is the square root of 4 squared plus 3 squared, which is 5xyA(1, 1)B(5, 4)Δx = 4Δy = 3AB = √(4² + 3²) = 5

The distance formula: Pythagoras on a grid

Two points joined by a segment, with a dashed right triangle under it. The horizontal leg is x2 minus x1 and the vertical leg is y2 minus y1, so the distance squared is the sum of their squaresxy(x₁, y₁)(x₂, y₂)x₂ − x₁y₂ − y₁d² = (x₂ − x₁)² + (y₂ − y₁)²
The horizontal and vertical changes are the legs of a right triangle. The segment is its hypotenuse.

To find the distance between (x₁, y₁) and (x₂, y₂), draw a right triangle under the segment. Its legs are the horizontal change Δx = x₂ − x₁ and the vertical change Δy = y₂ − y₁. By the Pythagorean theorem:

d = √((x₂ − x₁)² + (y₂ − y₁)²)

Example. From (−2, 1) to (4, −2): Δx = 6, Δy = −3, so d = √(36 + 9) = √45 = √(9 · 5) = 3√5 ≈ 6.71. The square makes a negative change positive, so the order of the points does not matter.

Exact answers. Leave a square root unless it comes out whole, and simplify it by taking out square factors: √72 = √(36 · 2) = 6√2.

The midpoint formula

Segment from A at 1, 1 to B at 9, 5. Its midpoint M at 5, 3 is found by averaging the x coordinates, 1 and 9, and the y coordinates, 1 and 5xyA(1, 1)B(9, 5)M(5, 3)M = ((1 + 9)/2, (1 + 5)/2)
The midpoint is halfway across and halfway up.

The midpoint of a segment averages the coordinates of its endpoints:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Working backwards. If M is the midpoint and you know one endpoint A, the other endpoint is B = (2xM − xA, 2yM − yA). Whatever step takes you from A to M takes you from M to B.

Slope, parallel and perpendicular

Two parallel lines, each with slope one half, and a third line with slope negative 2 crossing them at right angles. Parallel lines have equal slopes; perpendicular slopes multiply to negative 1xym = 1/2m = 1/2m = −2parallel: equal slopes · perpendicular: product −1
Two parallel lines with slope 1/2, and a perpendicular line with slope −2.

The slope of the line through two points is m = (y₂ − y₁)/(x₂ − x₁), rise over run. A horizontal line has slope 0; a vertical line has an undefined slope, because its run is 0. (Graphing lines from their equations is covered in Algebra: Graphing Linear Equations.)

  • Parallel lines have equal slopes (or are both vertical).
  • Perpendicular lines have slopes whose product is −1, so each is the negative reciprocal of the other: 2/3 and −3/2. A horizontal line is perpendicular to a vertical one.

Why −1? Turn the slope triangle with run 3 and rise 2 through 90°. The run and rise swap places and one changes sign: run −2, rise 3, so the new slope is 3/(−2) = −3/2.

Explorer: points on a grid

Drag the points, or type coordinates in the boxes (use the arrow keys in a box to step by 1). The readout shows the length, midpoint and slope of AB live. Turn on point C to test a triangle for a right angle.

Interactive coordinate grid with an x axis and a y axis, each from −6 to 6, with points A and B, their segment, its midpoint and a dashed slope triangle, and an optional point C forming a triangle
Point A
Point B

The equation of a circle

A circle with center 4, 3 and radius 5. The point 7, 7 is on it because its horizontal and vertical distances from the center, 3 and 4, satisfy 3 squared plus 4 squared equals 5 squaredxy(h, k) = (4, 3)(x, y)34r = 5(x − 4)² + (y − 3)² = 25
Every point on the circle is 5 from the center (4, 3).

A circle is every point at distance r from its center. Put the center at (h, k); a point (x, y) is on the circle exactly when the distance formula gives r. Squaring both sides:

(x − h)² + (y − k)² = r²

  • The center of (x + 2)² + (y − 5)² = 49 is (−2, 5), because x + 2 = x − (−2). The radius is √49 = 7, not 49.
  • A circle centered at the origin is x² + y² = r².
  • To test a point, substitute it: (7, 7) is on the circle in the figure because (7 − 4)² + (7 − 3)² = 9 + 16 = 25.

Coordinate proofs

Quadrilateral A 1, 1, B 5, 2, C 4, 6, D 0, 5. Opposite sides have slopes one quarter and negative 4, which multiply to negative 1, and all four sides have length root 17, so ABCD is a squarexyA(1, 1)B(5, 2)C(4, 6)D(0, 5)slopes:AB, DC: 1/4BC, AD: −4sides all √17so: a square
Slopes show the right angles; lengths show the equal sides.

In a coordinate proof you show a fact about a figure by computing with its coordinates. The three tools do different jobs:

  • Slope tests parallel (equal slopes) and perpendicular (product −1).
  • Distance tests equal lengths.
  • Midpoint tests whether segments bisect each other: a quadrilateral is a parallelogram exactly when its diagonals share a midpoint.

Classifying a quadrilateral. For A(1, 1), B(5, 2), C(4, 6), D(0, 5): slopes AB = DC = 1/4 and BC = AD = −4, so both pairs of opposite sides are parallel (a parallelogram), and (1/4)(−4) = −1, so the corners are right angles (a rectangle). All four sides are √(4² + 1²) = √17, so ABCD is a square. Name the most specific shape the evidence proves.

Showing a triangle is right. For P(0, 0), Q(4, 2), R(3, 4): slope PQ = 1/2 and slope QR = 2/(−1) = −2, and (1/2)(−2) = −1, so the right angle is at Q. As a check with the converse of the Pythagorean theorem: PQ² + QR² = 20 + 5 = 25 = PR².

Choosing coordinates. To prove something about every figure of a kind, place it conveniently and use letters: a rectangle at (0, 0), (a, 0), (a, b), (0, b). Its diagonals both have midpoint (a/2, b/2), so the diagonals of every rectangle bisect each other.

Common mistakes

  • Mixing up the order in slope. Subtract in the same order top and bottom: (y₂ − y₁)/(x₂ − x₁), never (y₂ − y₁)/(x₁ − x₂).
  • Run over rise. Slope is change in y over change in x.
  • Sign errors in a circle’s center. (x − 3)² + (y + 1)² = 16 has center (3, −1).
  • Forgetting the square root. In a circle’s equation the right side is r², and in the distance formula √(Δx² + Δy²) is not Δx + Δy.
  • Perpendicular slopes. The negative reciprocal of 3 is −1/3, not −3 and not 1/3.
  • Stopping too early. A rectangle with four equal sides is a square; give the most specific name.

History: Descartes, Fermat and the marriage of algebra and geometry

In 1637 René Descartes published his Discours de la méthode at Leiden with three appendices, one of them La Géométrie. In it he turned geometric problems into equations about lengths and solved them with algebra. That is why coordinates are still called Cartesian, after the Latin form of his name, Cartesius. His axes were not quite ours: the habit of drawing two perpendicular number lines with negative values came later.

At about the same time, Pierre de Fermat, a lawyer in Toulouse, reached the same idea on his own. Working from Apollonius’ lost book on Plane Loci, he wrote Ad locos planos et solidos isagoge (Introduction to Plane and Solid Loci), which showed that an equation in two unknowns describes a line or curve. He sent it to mathematicians in Paris, but never polished it for print, so Descartes got the public credit.

Self-guided quiz: Plot Twist

This self-guided quiz mixes distance, midpoints, slope, parallel and perpendicular lines, circles and coordinate proofs. Give exact answers: type a square root as √13 or sqrt 13, and 3√2 as 3√2 or 3 sqrt 2. Type points as (x, y).

Sources

Credits: all diagrams on this page are original SVG drawings made for Jugami.