Coordinate Geometry

Coordinate geometry lets us describe geometric figures using numbers. We draw two perpendicular number lines: the horizontal x-axis and the vertical y-axis. They meet at the origin O(0, 0).

A point P is written as an ordered pair (x, y). Here x is the abscissa (perpendicular distance from the y-axis) and y is the ordinate (perpendicular distance from the x-axis). Order matters: (3, 5) and (5, 3) are different points.

The two axes split the plane into four quadrants, and the sign pattern of (x, y) tells you the quadrant:

• Quadrant I: (+, +)
• Quadrant II: (−, +)
• Quadrant III: (−, −)
• Quadrant IV: (+, −)

Points on the x-axis have y = 0, e.g. (4, 0); points on the y-axis have x = 0, e.g. (0, −7).

Worked example: Where does A(−3, 2) lie? The abscissa is negative and the ordinate is positive, so the sign pattern is (−, +). Hence A lies in Quadrant II. Similarly B(5, 0) has ordinate 0, so B lies on the x-axis.

To find the distance between P(x₁, y₁) and Q(x₂, y₂), drop perpendiculars to form a right triangle. The horizontal side has length (x₂ − x₁) and the vertical side has length (y₂ − y₁). By Pythagoras' theorem, the hypotenuse PQ satisfies PQ² = (x₂ − x₁)² + (y₂ − y₁)².

Taking the positive square root (distance is never negative):

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

The distance of a point P(x, y) from the origin is a special case (take the other point as O(0, 0)): OP = √(x² + y²).

Worked example: Find the distance between P(2, 3) and Q(4, 1).

PQ = √[(4 − 2)² + (1 − 3)²] = √[(2)² + (−2)²] = √[4 + 4] = √8 = 2√2 units.

Worked example (distance from origin): The distance of (−4, 3) from the origin is √[(−4)² + 3²] = √[16 + 9] = √25 = 5 units.

Note: since we square the differences, it does not matter which point you call “first” — (x₂ − x₁)² equals (x₁ − x₂)².

Suppose a point P divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n (so AP : PB = m : n). Then the coordinates of P are given by the section formula:

P = ( (mx₂ + nx₁) / (m + n) , (my₂ + ny₁) / (m + n) )

A memory aid: B's coordinate is multiplied by m and A's coordinate by n — the multipliers are cross-associated with the ratio AP : PB = m : n (a larger m pulls P closer to B).

Worked example (finding the point): Find the point dividing the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3.

Here m = 2, n = 3. So x = (2·4 + 3·(−1))/(2+3) = (8 − 3)/5 = 1, and y = (2·(−3) + 3·7)/(2+3) = (−6 + 21)/5 = 15/5 = 3. Hence P = (1, 3).

Worked example (finding the ratio): In what ratio does P(−4, 6) divide the join of A(−6, 10) and B(3, −8)?

Let the ratio be k : 1. Using the x-coordinate: −4 = (3k + (−6))/(k + 1). So −4(k + 1) = 3k − 6, giving −4k − 4 = 3k − 6, hence 2 = 7k, so k = 2/7. The ratio is 2 : 7. (Check with y: (−8·2/7 + 10)/(2/7 + 1) = (54/7)/(9/7) = 6 ✓.)

The midpoint M of the segment joining A(x₁, y₁) and B(x₂, y₂) divides it in the ratio 1 : 1. Putting m = n = 1 in the section formula gives the simple average of coordinates:

M = ( (x₁ + x₂) / 2 , (y₁ + y₂) / 2 )

Worked example: The midpoint of A(6, −5) and B(−2, 11) is ((6 + (−2))/2, (−5 + 11)/2) = (4/2, 6/2) = (2, 3).

Worked example (finding an endpoint): If M(1, 4) is the midpoint of the segment joining A(−3, 6) and B(x, y), find B.

Using (−3 + x)/2 = 1 gives −3 + x = 2, so x = 5. Using (6 + y)/2 = 4 gives 6 + y = 8, so y = 2. Hence B = (5, 2).

The midpoint idea is powerful for quadrilaterals: the diagonals of a parallelogram bisect each other, so the midpoint of one diagonal equals the midpoint of the other. This lets you find a missing vertex.

Collinearity. Three points A, B, C are collinear (lie on one straight line) if the longest of the three distances equals the sum of the other two, for example AB + BC = AC.

Worked example: Are A(1, −1), B(5, 2), C(9, 5) collinear? AB = √[(5−1)² + (2+1)²] = √[16 + 9] = 5. BC = √[(9−5)² + (5−2)²] = √[16 + 9] = 5. AC = √[(9−1)² + (5+1)²] = √[64 + 36] = √100 = 10. Since AB + BC = 5 + 5 = 10 = AC, the points are collinear.

Classifying a quadrilateral. For vertices in order, compute all four sides and both diagonals, then decide:

• Square: all four sides equal AND diagonals equal.
• Rhombus: all four sides equal BUT diagonals unequal.
• Rectangle: opposite sides equal AND diagonals equal.
• Parallelogram: opposite sides equal (equivalently, diagonals bisect each other).

Worked example: Show A(1, 7), B(4, 2), C(−1, −1), D(−4, 4) form a square.

AB = √[(4−1)² + (2−7)²] = √[9 + 25] = √34. BC = √[(−1−4)² + (−1−2)²] = √[25 + 9] = √34. CD = √[(−4+1)² + (4+1)²] = √[9 + 25] = √34. DA = √[(1+4)² + (7−4)²] = √[25 + 9] = √34. Diagonals: AC = √[(−1−1)² + (−1−7)²] = √[4 + 64] = √68 = 2√17, and BD = √[(−4−4)² + (4−2)²] = √[64 + 4] = √68 = 2√17. All four sides are equal and both diagonals are equal, so ABCD is a square.