Chapter 06 — Coordinate Geometry | Part 1: Straight Lines, Gradients and Equations

MATHEMATICS HIPHOP ACADEMY

CHAPTER 06

COORDINATE GEOMETRY

PART 1: STRAIGHT LINES, GRADIENTS AND EQUATIONS

A-Level Pure Mathematics • Detailed Notes • Worked Examples • Exercises

Learning Objectives

By the end of this lesson, you should be able to:

• Understand Cartesian coordinates and ordered pairs.
• Calculate the distance between two points.
• Find the midpoint of a line segment.
• Calculate the gradient of a straight line.
• Understand the relationship between gradient and angle.
• Form equations of straight lines using different methods.
• Determine the intercepts of a straight line.
• Find points of intersection of two straight lines.
• Apply the conditions for parallel and perpendicular lines.
• Calculate the perpendicular distance from a point to a line.

1. Introduction to Coordinate Geometry

Coordinate geometry is the branch of mathematics that combines algebra and geometry. It allows us to describe geometrical figures using numbers, coordinates and equations.

In coordinate geometry, points are located on a flat surface called the Cartesian plane, which consists of two perpendicular number lines:

• x-axis: the horizontal number line.
• y-axis: the vertical number line.
• Origin: the point where the two axes meet, represented by (0, 0).

The Cartesian plane showing the two axes and a point P.

1.1 Ordered Pairs

A point on the Cartesian plane is represented by an ordered pair (x, y).

The first coordinate gives the horizontal position, while the second gives the vertical position.

For example, the point A(3, 2) means:

• Move 3 units to the right from the origin.
• Move 2 units upwards.

The order is important. A(3, 2) and B(2, 3) represent different points.

2. Distance Between Two Points

The distance between two points is the length of the straight line joining them.

Consider two points:

A(x₁, y₁) and B(x₂, y₂)

Using Pythagoras' theorem, the horizontal difference is (x₂ − x₁), while the vertical difference is (y₂ − y₁).

DISTANCE FORMULA

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

Worked Example 1

Find the distance between A(2, 3) and B(8, 11).

Solution

Using:

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

Substituting the coordinates:

AB = √[(8 − 2)² + (11 − 3)²]

AB = √[6² + 8²]

AB = √(36 + 64)

AB = √100

AB = 10 units.

Worked Example 2

The distance between P(−2, 4) and Q(a, 1) is 5 units. Find the possible values of a.

Solution

Using the distance formula:

5² = (a − (−2))² + (1 − 4)²

25 = (a + 2)² + 9

(a + 2)² = 16

a + 2 = ±4

a = 2 or a = −6.

3. Midpoint of a Line Segment

The midpoint of a line segment is the point that divides the segment into two equal parts.

If A(x₁, y₁) and B(x₂, y₂) are the endpoints of a line segment, its midpoint M is given by:

MIDPOINT FORMULA

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Worked Example 3

Find the midpoint of the line joining A(−4, 6) and B(8, 2).

Solution

M = ((−4 + 8)/2, (6 + 2)/2)

M = (4/2, 8/2)

M = (2, 4).

Worked Example 4

The midpoint of A(3, 5) and B(x, y) is M(7, 2). Find x and y.

Solution

Using the midpoint formula:

(3 + x)/2 = 7

3 + x = 14

x = 11

Also:

(5 + y)/2 = 2

5 + y = 4

y = −1

Therefore, B = (11, −1).

4. Gradient of a Straight Line

The gradient of a straight line measures its steepness and direction.

It is defined as the ratio of the vertical change to the horizontal change between two points on the line.

GRADIENT FORMULA

m = (y₂ − y₁)/(x₂ − x₁)

Change in y divided by change in x.

4.1 Types of Gradients

• Positive gradient: the line rises from left to right.
• Negative gradient: the line falls from left to right.
• Zero gradient: the line is horizontal.
• Undefined gradient: the line is vertical because the change in x is zero.

Worked Example 5

Find the gradient of the line joining A(2, 3) and B(6, 11).

Solution

m = (11 − 3)/(6 − 2)

m = 8/4

m = 2.

Worked Example 6

Find the gradient of the line joining P(−3, 7) and Q(5, −1).

Solution

m = (−1 − 7)/(5 − (−3))

m = −8/8

m = −1.

5. Gradient and the Angle of Inclination

The angle of inclination of a straight line is the angle θ measured anticlockwise from the positive x-axis to the line.

The gradient of a line is related to its angle of inclination by:

GRADIENT-ANGLE RELATIONSHIP

m = tan θ

For example, if θ = 45°:

m = tan 45° = 1.

If θ = 135°:

m = tan 135° = −1.

Important: A vertical line has an inclination of 90°, but its gradient is undefined because tan 90° is undefined.

Worked Example 7

A straight line has an angle of inclination of 30°. Find its gradient.

Solution

m = tan θ

m = tan 30°

m = 1/√3.

6. Equation of a Straight Line

The equation of a straight line represents the relationship between the x-coordinates and y-coordinates of all points lying on that line.

6.1 Slope-Intercept Form

The most common form is:

y = mx + c

m = gradient; c = y-intercept.

In this equation:

• m represents the gradient.
• c represents the value of y when x = 0.
• The line crosses the y-axis at (0, c).

Worked Example 8

Find the gradient and y-intercept of the line:

y = 3x − 5

Solution

Comparing with y = mx + c:

m = 3 and c = −5.

Gradient = 3; y-intercept = −5.

6.2 Point-Gradient Form

If a straight line passes through the point (x₁, y₁) and has gradient m, its equation is:

y − y₁ = m(x − x₁)

Worked Example 9

Find the equation of a straight line passing through (2, 5) with gradient 3.

Solution

Using y − y₁ = m(x − x₁):

y − 5 = 3(x − 2)

y − 5 = 3x − 6

y = 3x − 1.

6.3 Equation Given Two Points

When two points on a line are known, first calculate the gradient, then substitute one of the points into the point-gradient equation.

Worked Example 10

Find the equation of the line passing through A(1, 4) and B(3, 10).

Solution

First calculate the gradient:

m = (10 − 4)/(3 − 1)

m = 6/2 = 3.

Using point A(1, 4):

y − 4 = 3(x − 1)

y − 4 = 3x − 3

y = 3x + 1.

6.4 General Form of a Straight Line

The general Cartesian equation of a straight line is:

ax + by + c = 0

a, b and c are constants, with a and b not both zero.

When b is not zero, the equation can be rearranged to make y the subject:

by = −ax − c

y = (−a/b)x − c/b

Therefore, the gradient is:

m = −a/b.

Worked Example 11

Find the gradient of the line:

3x + 2y − 8 = 0

Solution

Rearranging:

2y = −3x + 8

y = −(3/2)x + 4

Gradient = −3/2.

6.5 Intercept Form

If a line cuts the x-axis at (a, 0) and the y-axis at (0, b), its equation may be written as:

x/a + y/b = 1

This form is useful when the two intercepts are known and neither intercept is zero.

Worked Example 12

A straight line cuts the x-axis at 4 and the y-axis at 6. Find its equation.

Solution

Using the intercept form:

x/4 + y/6 = 1

Multiplying by 12:

3x + 2y = 12

3x + 2y − 12 = 0.

7. Intercepts of a Straight Line

An intercept is the point where a straight line crosses one of the coordinate axes.

7.1 x-intercept

At the x-axis, y = 0. Therefore, substitute y = 0 into the equation.

7.2 y-intercept

At the y-axis, x = 0. Therefore, substitute x = 0 into the equation.

Worked Example 13

Find the x-intercept and y-intercept of:

2x + 3y = 12

Solution

For the x-intercept:

Put y = 0:

2x = 12, hence x = 6.

x-intercept = (6, 0).

For the y-intercept:

Put x = 0:

3y = 12, hence y = 4.

y-intercept = (0, 4).

8. Point of Intersection of Two Straight Lines

The point of intersection is the point that lies on both lines. Its coordinates satisfy both equations simultaneously.

To find the intersection, solve the two equations simultaneously using substitution or elimination.

Worked Example 14

Find the point of intersection of:

y = 2x + 1

y = −x + 7

Solution

Since both expressions equal y:

2x + 1 = −x + 7

3x = 6

x = 2.

Substitute into the first equation:

y = 2(2) + 1 = 5.

Point of intersection = (2, 5).

9. Parallel Straight Lines

Parallel lines are lines in the same plane that never meet, however far they are extended.

Two non-vertical straight lines are parallel if and only if they have equal gradients.

CONDITION FOR PARALLEL LINES

m₁ = m₂

For example:

y = 3x + 2 and y = 3x − 5 are parallel because both have gradient 3.

Worked Example 15

Find the equation of a line parallel to y = 2x + 3 and passing through (4, 1).

Solution

The given line has gradient 2.

A parallel line has the same gradient.

Using point-gradient form:

y − 1 = 2(x − 4)

y − 1 = 2x − 8

y = 2x − 7.

10. Perpendicular Straight Lines

Perpendicular lines meet at a right angle of 90°.

If two non-vertical straight lines have gradients m₁ and m₂, they are perpendicular when:

CONDITION FOR PERPENDICULAR LINES

m₁ × m₂ = −1

Equivalently, m₂ = −1/m₁ when m₁ is non-zero.

The gradients of two perpendicular non-vertical lines are negative reciprocals of each other.

Worked Example 16

Find the gradient of a line perpendicular to a line whose gradient is 4.

Solution

Let the required gradient be m.

4 × m = −1

m = −1/4.

Worked Example 17

Find the equation of a line perpendicular to y = −2x + 5 and passing through (2, 3).

Solution

The given gradient is −2.

The perpendicular gradient is:

m = 1/2.

Using point-gradient form:

y − 3 = (1/2)(x − 2)

y − 3 = x/2 − 1

y = x/2 + 2.

11. Perpendicular Distance from a Point to a Line

The perpendicular distance from a point to a straight line is the shortest distance between the point and the line.

For a line written in the general form:

ax + by + c = 0

and a point P(x₁, y₁), the perpendicular distance d is:

PERPENDICULAR DISTANCE FORMULA

d = |ax₁ + by₁ + c| / √(a² + b²)

Worked Example 18

Find the perpendicular distance from P(2, 1) to the line 3x + 4y − 10 = 0.

Solution

Here, a = 3, b = 4, c = −10, x₁ = 2 and y₁ = 1.

Using the formula:

d = |3(2) + 4(1) − 10| / √(3² + 4²)

d = |6 + 4 − 10| / √25

d = 0/5

d = 0 units.

Interpretation: The point P lies on the line itself.

Worked Example 19

Find the perpendicular distance from P(1, 2) to the line 3x + 4y + 5 = 0.

Solution

d = |3(1) + 4(2) + 5| / √(3² + 4²)

d = |3 + 8 + 5| / √25

d = 16/5

d = 3.2 units.

12. Summary of Important Formulae

Distance: d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Midpoint: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Gradient: m = (y₂ − y₁)/(x₂ − x₁)

Angle: m = tan θ

Slope-intercept: y = mx + c

Point-gradient: y − y₁ = m(x − x₁)

General equation: ax + by + c = 0

Parallel lines: m₁ = m₂

Perpendicular lines: m₁m₂ = −1

Point-to-line distance: d = |ax₁ + by₁ + c|/√(a² + b²)

13. Examination Practice

Attempt all the following questions before opening the solutions. Show your working clearly.

Exercise 1: Basic Coordinate Geometry

1. Find the distance between A(−2, 3) and B(4, 11).
2. Find the midpoint of the line joining P(−5, 7) and Q(3, −1).
3. The midpoint of A(2, 4) and B(x, y) is (5, 1). Find x and y.
4. Find the gradient of the line joining (−3, 2) and (5, 6).
5. Find the gradient of a line whose angle of inclination is 150°.

Exercise 2: Equations of Straight Lines

6. Find the equation of a line with gradient 4 passing through (2, 3).
7. Find the equation of the line passing through A(1, 2) and B(3, 8).
8. Find the gradient and y-intercept of 5x − 2y + 6 = 0.
9. Find the x-intercept and y-intercept of 3x + 2y = 12.
10. A line passes through (−2, 5) and has gradient −3. Express its equation in general form.

Exercise 3: Parallel and Perpendicular Lines

11. Find the gradient of a line parallel to y = −4x + 7.
12. Find the gradient of a line perpendicular to y = (1/3)x − 2.
13. Find the equation of a line parallel to 2x − 3y + 6 = 0 and passing through (3, 1).
14. Find the equation of a line perpendicular to y = 2x + 4 and passing through (4, 1).

Exercise 4: Further Applications

15. Find the point of intersection of y = 3x − 2 and y = −2x + 8.
16. Find the perpendicular distance from (2, 3) to the line 3x + 4y − 7 = 0.
17. The line joining A(1, 2) and B(5, k) has gradient 3. Find k.
18. Find the equation of the line passing through the midpoint of A(2, 6) and B(8, 4), and having gradient −2.
19. Find the value of p if the lines 3x + py − 5 = 0 and 2x − 4y + 7 = 0 are perpendicular.

14. Worked Exercise Solutions

Open each solution only after attempting the corresponding question.

Solutions 1–5

1. d = √[(4 + 2)² + (11 − 3)²] = √(36 + 64) = 10 units.

2. M = ((−5 + 3)/2, (7 − 1)/2) = (−1, 3).

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

4. m = (6 − 2)/(5 + 3) = 4/8 = 1/2.

5. m = tan 150° = −1/√3.

Solutions 6–10

6. y − 3 = 4(x − 2), hence y = 4x − 5.

7. m = (8 − 2)/(3 − 1) = 3. Therefore y − 2 = 3(x − 1), giving y = 3x − 1.

8. 5x − 2y + 6 = 0 gives y = (5/2)x + 3. Gradient = 5/2; y-intercept = 3.

9. Put y = 0: x = 4. Put x = 0: y = 6. Intercepts are (4, 0) and (0, 6).

10. y − 5 = −3(x + 2). Hence y = −3x − 1, or 3x + y + 1 = 0.

Solutions 11–14

11. Parallel lines have equal gradients. Required gradient = −4.

12. The negative reciprocal of 1/3 is −3.

13. 2x − 3y + 6 = 0 gives gradient 2/3. Thus y − 1 = (2/3)(x − 3). Hence 2x − 3y − 3 = 0.

14. Perpendicular gradient = −1/2. Thus y − 1 = −(1/2)(x − 4), giving x + 2y − 6 = 0.

Solutions 15–19

15. 3x − 2 = −2x + 8, so 5x = 10 and x = 2. Then y = 4. Intersection = (2, 4).

16. d = |3(2) + 4(3) − 7|/√(9 + 16) = 11/5. Distance = 11/5 units.

17. (k − 2)/(5 − 1) = 3. Hence k − 2 = 12, giving k = 14.

18. Midpoint = ((2 + 8)/2, (6 + 4)/2) = (5, 5). Equation: y − 5 = −2(x − 5), giving 2x + y − 15 = 0.

19. First gradient = −3/p; second gradient = 1/2. For perpendicular lines, (−3/p)(1/2) = −1. Hence p = 3/2.

End of Part 1

You should now be able to calculate distances and midpoints, determine gradients, form equations of straight lines, find intersections, and apply the conditions for parallel and perpendicular lines.

Make sure you practise rearranging equations and showing every step of your working. These skills will be important in the next parts of Coordinate Geometry.

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