Coordinate Geometry

What you'll learn

Coordinate geometry connects algebra with geometry by using coordinate systems to describe positions, shapes and relationships. This topic is essential for CSEC Additional Mathematics and accounts for approximately 10-15% of Paper 02. You'll master equations of straight lines, calculate distances and midpoints, determine gradients, and solve problems involving parallel and perpendicular lines.

Key terms and definitions

Gradient (m) — the measure of steepness of a line, calculated as the ratio of vertical change to horizontal change between two points; a positive gradient slopes upward from left to right, while a negative gradient slopes downward.

y-intercept (c) — the y-coordinate of the point where a line crosses the y-axis; it occurs when x = 0.

Collinear points — three or more points that lie on the same straight line; verified when the gradient between any two pairs of points is identical.

Perpendicular lines — two lines that intersect at a right angle (90°); their gradients multiply to give -1, expressed as m₁ × m₂ = -1.

Parallel lines — lines that never intersect and maintain constant separation; they have identical gradients but different y-intercepts.

Distance formula — the expression used to calculate the straight-line distance between two points in a coordinate plane, derived from Pythagoras' theorem.

Midpoint — the point that divides a line segment into two equal parts, located exactly halfway between two endpoints.

Equation of a line — a mathematical statement showing the relationship between x and y coordinates for all points on that line, commonly written in forms such as y = mx + c or ax + by + c = 0.

Core concepts

Distance between two points

The distance formula calculates the length of a line segment joining two points A(x₁, y₁) and B(x₂, y₂):

$$d = \sqrt{(x₂ - x₁)² + (y₂ - y₁)²}$$

This formula comes directly from Pythagoras' theorem. The horizontal distance (x₂ - x₁) and vertical distance (y₂ - y₁) form the two shorter sides of a right-angled triangle, with the distance d as the hypotenuse.

Key applications:

• Finding the length of line segments
• Verifying if a triangle is isosceles, equilateral or scalene
• Determining if points form a specific quadrilateral
• Calculating perimeters of polygons

Important note: Always square the differences before adding them. The order of subtraction doesn't matter because squaring eliminates negative signs.

Midpoint of a line segment

The midpoint M of a line segment joining A(x₁, y₁) and B(x₂, y₂) has coordinates:

$$M = \left(\frac{x₁ + x₂}{2}, \frac{y₁ + y₂}{2}\right)$$

You find the midpoint by averaging the x-coordinates and averaging the y-coordinates separately.

Key applications:

• Finding the centre of a line segment
• Locating the point of intersection of diagonals in parallelograms and rectangles
• Solving problems involving bisectors

If you know the midpoint M(x, y) and one endpoint A(x₁, y₁), you can find the other endpoint B(x₂, y₂) by rearranging:

• x₂ = 2x - x₁
• y₂ = 2y - y₁

Gradient and equation of a straight line

The gradient (or slope) between two points A(x₁, y₁) and B(x₂, y₂) is:

$$m = \frac{y₂ - y₁}{x₂ - x₁} = \frac{\text{vertical change}}{\text{horizontal change}} = \frac{\text{rise}}{\text{run}}$$

Properties of gradients:

• Positive gradient: line slopes upward from left to right
• Negative gradient: line slopes downward from left to right
• Zero gradient: horizontal line (y = constant)
• Undefined gradient: vertical line (x = constant)

Forms of linear equations:

1. Gradient-intercept form: y = mx + c

   • m is the gradient
   • c is the y-intercept
   • Most useful form for quick graphing

2. Point-gradient form: y - y₁ = m(x - x₁)

   • Used when you know a point (x₁, y₁) and gradient m
   • Rearrange to gradient-intercept form for final answers

3. General form: ax + by + c = 0

   • Standard form for CXC examinations
   • All terms on one side, equation equal to zero
   • Coefficients should be integers with no common factors

4. Two-point form: When given two points, find gradient first, then use point-gradient form with either point.

Parallel and perpendicular lines

Parallel lines:

• Have equal gradients: m₁ = m₂
• Never intersect
• Distance between them remains constant

If line L₁ has equation y = m₁x + c₁ and line L₂ is parallel to L₁, then L₂ has equation y = m₁x + c₂ where c₂ ≠ c₁.

Perpendicular lines:

• Their gradients multiply to give -1: m₁ × m₂ = -1
• Therefore, m₂ = -1/m₁ (negative reciprocal)
• They intersect at 90°

Examples of perpendicular gradients:

• If m₁ = 2, then m₂ = -1/2
• If m₁ = -3/4, then m₂ = 4/3
• If m₁ = 5, then m₂ = -1/5

Special cases:

• Horizontal line (m = 0) is perpendicular to vertical line (m undefined)
• Lines y = k and x = h are always perpendicular

Collinearity and geometric applications

Three points A, B, and C are collinear if they lie on the same straight line. To verify collinearity:

Method 1 — Gradient comparison: Calculate gradient of AB and gradient of BC. If equal, points are collinear.

Method 2 — Distance check: Calculate AB + BC and AC. If AB + BC = AC, the points are collinear with B between A and C.

Geometric shapes:

Use coordinate geometry to prove properties of triangles and quadrilaterals:

• Parallelogram: Opposite sides parallel (equal gradients) or diagonals bisect each other (same midpoint)
• Rectangle: Parallelogram with adjacent sides perpendicular
• Rhombus: Parallelogram with all sides equal length
• Square: Rectangle with all sides equal length
• Isosceles triangle: Two sides of equal length
• Right-angled triangle: Two sides perpendicular (gradients multiply to -1)

Intersection of lines

To find where two lines intersect, solve their equations simultaneously.

Method:

1. Write both equations clearly
2. Use substitution or elimination
3. Solve for one variable
4. Substitute back to find the other variable
5. Express answer as coordinates (x, y)

If lines are parallel, there is no intersection point (inconsistent system). The attempt to solve will lead to a contradiction like 0 = 5.

Worked examples

Example 1: Distance, midpoint and gradient

The coordinates of two ports in the Caribbean are A(2, 5) and B(8, 13), where units are in kilometres.

(a) Calculate the distance AB.

(b) Find the coordinates of the midpoint M of AB.

(c) Determine the gradient of AB.

(d) A shipping lane from port C(-4, 9) to port M is perpendicular to AB. Find the equation of line CM.

Solution:

(a) Using the distance formula: $$d = \sqrt{(8 - 2)² + (13 - 5)²}$$ $$d = \sqrt{6² + 8²}$$ $$d = \sqrt{36 + 64}$$ $$d = \sqrt{100}$$ $$d = 10 \text{ km}$$

(b) Using the midpoint formula: $$M = \left(\frac{2 + 8}{2}, \frac{5 + 13}{2}\right)$$ $$M = \left(\frac{10}{2}, \frac{18}{2}\right)$$ $$M = (5, 9)$$

(c) Using the gradient formula: $$m = \frac{13 - 5}{8 - 2} = \frac{8}{6} = \frac{4}{3}$$

(d) Since CM is perpendicular to AB: $$m_{CM} = -\frac{1}{m_{AB}} = -\frac{1}{\frac{4}{3}} = -\frac{3}{4}$$

Using point-gradient form with C(-4, 9): $$y - 9 = -\frac{3}{4}(x - (-4))$$ $$y - 9 = -\frac{3}{4}(x + 4)$$ $$y - 9 = -\frac{3}{4}x - 3$$ $$y = -\frac{3}{4}x + 6$$

Converting to general form: $$4y = -3x + 24$$ $$3x + 4y - 24 = 0$$

Example 2: Proving geometric properties

The vertices of a quadrilateral are P(1, 2), Q(4, 5), R(7, 2) and S(4, -1).

(a) Show that PQRS is a rhombus.

(b) Prove that the diagonals PR and QS are perpendicular.

Solution:

(a) To prove PQRS is a rhombus, show all four sides are equal length.

$$PQ = \sqrt{(4-1)² + (5-2)²} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$

$$QR = \sqrt{(7-4)² + (2-5)²} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$

$$RS = \sqrt{(4-7)² + (-1-2)²} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$

$$SP = \sqrt{(1-4)² + (2-(-1))²} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$

Since PQ = QR = RS = SP, all sides are equal, therefore PQRS is a rhombus.

(b) Find gradients of diagonals PR and QS:

$$m_{PR} = \frac{2-2}{7-1} = \frac{0}{6} = 0$$

$$m_{QS} = \frac{-1-5}{4-4} = \frac{-6}{0}$$ (undefined — vertical line)

A horizontal line (m = 0) is perpendicular to a vertical line (m undefined), therefore the diagonals are perpendicular.

Example 3: Line through intersection

Line L₁ has equation 2x + y - 8 = 0 and line L₂ has equation x - y - 1 = 0.

(a) Find the coordinates of the point of intersection of L₁ and L₂.

(b) A sugar cane plantation boundary runs along line L₃ which passes through the intersection point and is parallel to the line 3x + 2y - 5 = 0. Find the equation of L₃.

Solution:

(a) Solve simultaneously: $$2x + y = 8 \quad \text{...(1)}$$ $$x - y = 1 \quad \text{...(2)}$$

Adding equations (1) and (2): $$3x = 9$$ $$x = 3$$

Substituting x = 3 into equation (2): $$3 - y = 1$$ $$y = 2$$

Point of intersection is (3, 2).

(b) From 3x + 2y - 5 = 0, rearrange to find gradient: $$2y = -3x + 5$$ $$y = -\frac{3}{2}x + \frac{5}{2}$$

Gradient = -3/2

Since L₃ is parallel, it has the same gradient. Using point-gradient form with (3, 2): $$y - 2 = -\frac{3}{2}(x - 3)$$ $$y - 2 = -\frac{3}{2}x + \frac{9}{2}$$ $$y = -\frac{3}{2}x + \frac{13}{2}$$

Converting to general form: $$2y = -3x + 13$$ $$3x + 2y - 13 = 0$$

Common mistakes and how to avoid them

• Reversing coordinates in formulas: Always maintain the order (x, y). When working with two points, keep x-coordinates together and y-coordinates together. For gradient, ensure you subtract y-coordinates in the numerator and corresponding x-coordinates in the denominator.

• Incorrect perpendicular gradient: Remember m₁ × m₂ = -1, not m₁ + m₂ = 0. The perpendicular gradient is the negative reciprocal. If m = 2/3, the perpendicular gradient is -3/2, not -2/3.

• Forgetting to square before adding in distance formula: The formula is √[(x₂-x₁)² + (y₂-y₁)²], not √[(x₂-x₁) + (y₂-y₁)]². Square each difference separately before adding.

• Leaving equations in non-standard form: CXC expects final answers in the form ax + by + c = 0 where a, b, c are integers with no common factors. Convert from y = mx + c form and simplify fully.

• Not checking if gradient is undefined: When x₂ = x₁, the line is vertical with equation x = k (a constant). The gradient is undefined, not zero. Vertical lines have no y = mx + c form.

• Careless algebra when converting between forms: When multiplying through by denominators to clear fractions, multiply every term. When expanding brackets, apply the distributive property correctly. Check your work by substituting a known point.

Exam technique for "Coordinate Geometry"

• "Show that" questions require full working: When asked to "show" or "prove" a result, you must demonstrate every step clearly. Simply stating the answer, even if correct, earns zero marks. Calculate all distances, gradients or other required values explicitly.

• Present coordinates clearly: Always write coordinates as ordered pairs with parentheses: (3, 5), not 3, 5. When finding midpoints or intersection points, box or clearly identify your final coordinate answer.

• Identify command words and allocate time accordingly: "Calculate" requires numerical work (2-3 marks). "Find the equation" needs full algebraic manipulation to general form (3-4 marks). "Prove" or "show that" demands complete logical reasoning (4-5 marks). A 13-mark coordinate geometry question typically takes 15-18 minutes.

• Use exact values unless told otherwise: Leave answers with surds (√18 = 3√2) or fractions (m = 4/3) unless the question asks you to give the answer to a specific number of decimal places or significant figures. Only round at the final step.

Quick revision summary

Coordinate geometry links algebra and geometry through the Cartesian plane. Master the distance formula d = √[(x₂-x₁)² + (y₂-y₁)²] and midpoint formula M = ((x₁+x₂)/2, (y₁+y₂)/2). Calculate gradient as m = (y₂-y₁)/(x₂-x₁) and write line equations in form ax + by + c = 0. Parallel lines share equal gradients; perpendicular lines have gradients whose product equals -1. Prove collinearity by showing equal gradients between point pairs. Always show full working, present answers in required form, and verify results by substitution.