Coordinate geometry: gradient and straight-line equations

Overview

Coordinate geometry turns positions on a grid into numbers that can be measured and calculated. In this topic, a learner studies straight lines by finding their gradients and writing their equations. The gradient tells how quickly $y$ changes when $x$ changes, while the equation gives a compact rule for all points on the line.

This work matters because straight-line models appear in road design, maps, business graphs, science measurements, and later algebra. A line equation can describe a cost, a distance, a boundary, or a route, provided the relationship changes at a constant rate.

The most important habit in this chapter is to connect three views of the same line: the drawn graph, the coordinates of points on it, and the equation. A learner who can move between these views will find later graphing and simultaneous-equation topics much easier.

Prerequisites

• Reading coordinates as ordered pairs $(x,y)$.
• Plotting points on the Cartesian plane.
• Substituting values into simple algebraic expressions.
• Solving linear equations in one unknown.
• Understanding positive and negative numbers on a number line.

Learning Scope

This page covers gradients of straight lines, equations of straight lines, intercepts, simple graph interpretation, and the relationship between parallel and perpendicular lines. It focuses on linear relationships in two variables.

It does not cover midpoint and distance formulae in depth, which belong to Coordinate geometry: midpoint, distance, parallel lines, and perpendicular lines. It also does not replace the algebra topic Linear simultaneous equations or the graphing topic Graphical solution of simultaneous equations, although both use straight-line equations.

Subtopics

The Coordinate Plane

The Cartesian plane has a horizontal $x$-axis and a vertical $y$-axis. A point $(x,y)$ tells how far to move horizontally from the origin and then vertically. For example, $(3,2)$ is three units to the right and two units up from $(0,0)$.

Key insight: a straight line is a set of points that follow the same linear rule. If many points satisfy the same equation, they lie on the same line.

Coordinates are written in the order $(x,y)$. The $x$ value tells the horizontal movement first, and the $y$ value tells the vertical movement second. This order matters: $(3,2)$ and $(2,3)$ are different points.

Misconception note: a point is not named by the nearest axis only. To plot $(-4,1)$, move $4$ units left from the origin and then $1$ unit up.

Gradient As Rate Of Change

The gradient of a straight line measures steepness:

$$ m=\frac{\text{change in }y}{\text{change in }x} $$

For two points $(x_1,y_1)$ and $(x_2,y_2)$ on a non-vertical line,

$$ m=\frac{y_2-y_1}{x_2-x_1} $$

A positive gradient rises from left to right. A negative gradient falls from left to right. A zero gradient gives a horizontal line. A vertical line has undefined gradient because the change in $x$ is $0$.

The phrase "change in $y$ over change in $x$" can be remembered as:

$$ m=\frac{\text{rise}}{\text{run}} $$

If a line moves $3$ units up while moving $2$ units right, its gradient is $\frac{3}{2}$. If it moves $3$ units down while moving $2$ units right, its gradient is $-\frac{3}{2}$.

Misconception note: a steep line does not always have a positive gradient. Steepness tells size; direction tells sign.

Equations In Gradient-Intercept Form

The form

$$ y=mx+c $$

is useful because $m$ is the gradient and $c$ is the $y$-intercept. The $y$-intercept is the value of $y$ when $x=0$, so it is the point where the line crosses the $y$-axis.

Example: in $y=2x-3$, the gradient is $2$ and the $y$-intercept is $-3$. The line passes through $(0,-3)$ and rises $2$ units for every $1$ unit moved to the right.

To sketch from $y=mx+c$, begin at the intercept $(0,c)$, then use the gradient as a movement. For $y=2x-3$, start at $(0,-3)$, move $1$ unit right and $2$ units up to reach $(1,-1)$, then draw the line through the points.

Misconception note: the $c$ in $y=mx+c$ is not the constant "answer". It is the $y$ value when $x=0$.

Equations From A Point And A Gradient

If a line has gradient $m$ and passes through $(x_1,y_1)$, use the point-gradient form:

$$ y-y_1=m(x-x_1) $$

This form is especially helpful when the intercept is not known at the start. After substituting the point and gradient, the equation can be rearranged into $y=mx+c$ or into the general form $ax+by+c=0$.

The point-gradient form says: "Start from this known point, then move with this gradient." For a line of gradient $4$ through $(2,7)$:

$$ y-7=4(x-2) $$

Only after this substitution do you expand and simplify.

Equations From Two Points

When two points on a line are known, first find the gradient. Then use either point in the point-gradient form.

For points $(x_1,y_1)$ and $(x_2,y_2)$:

$$ \begin{aligned} m&=\frac{y_2-y_1}{x_2-x_1} \\ y-y_1&=m(x-x_1) \end{aligned} $$

The two points must be distinct. If $x_1=x_2$ and $y_1 \ne y_2$, the line is vertical and has equation $x=x_1$.

When using two points, keep the subtraction order consistent. If the numerator is $y_2-y_1$, the denominator must be $x_2-x_1$. Reversing both orders gives the same gradient, but reversing only one order changes the sign.

For example, from $(1,4)$ and $(5,12)$:

$$ \frac{12-4}{5-1}=\frac{8}{4}=2 $$

Also:

$$ \frac{4-12}{1-5}=\frac{-8}{-4}=2 $$

Parallel And Perpendicular Lines

Parallel non-vertical lines have equal gradients. If one line has gradient $m$, any line parallel to it also has gradient $m$.

Perpendicular lines meet at right angles. If two non-vertical lines are perpendicular, their gradients multiply to $-1$:

$$ m_1m_2=-1 $$

So if one gradient is $m_1$, the perpendicular gradient is

$$ m_2=-\frac{1}{m_1} $$

This relationship is often used when a question gives one line and asks for another line that passes through a given point.

Horizontal and vertical lines need special care. A horizontal line is perpendicular to a vertical line. The negative-reciprocal rule is used only when both gradients are defined and non-zero.

General Form Of A Straight Line

The general form is

$$ ax+by+c=0 $$

where $a$, $b$, and $c$ are constants and $a$ and $b$ are not both zero. To find the gradient from this form, rearrange to make $y$ the subject:

$$ \begin{aligned} ax+by+c&=0 \\ by&=-ax-c \\ y&=-\frac{a}{b}x-\frac{c}{b} \end{aligned} $$

For $b \ne 0$, the gradient is $-\frac{a}{b}$.

To find intercepts from general form, set one variable to $0$ at a time. For $2x+3y-12=0$:

• when $x=0$, $3y=12$, so $y=4$
• when $y=0$, $2x=12$, so $x=6$

The line crosses the axes at $(0,4)$ and $(6,0)$.

Key Terms

• Coordinate: an ordered pair $(x,y)$ showing position on the Cartesian plane.
• Cartesian plane: the grid formed by the perpendicular $x$-axis and $y$-axis.
• Gradient: the ratio of vertical change to horizontal change on a straight line.
• Intercept: a point where a graph crosses an axis.
• $y$-intercept: the value of $y$ when $x=0$.
• Linear equation: an equation whose graph is a straight line.
• Gradient-intercept form: the form $y=mx+c$.
• Point-gradient form: the form $y-y_1=m(x-x_1)$.
• General form: the form $ax+by+c=0$.
• Parallel lines: lines in the same plane that do not meet.
• Perpendicular lines: lines that meet at a right angle.

Worked Examples

Example 1: Find The Gradient From Two Points

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

Use the gradient formula:

$$ \begin{aligned} m&=\frac{y_2-y_1}{x_2-x_1} \\ &=\frac{11-3}{6-2} \\ &=\frac{8}{4} \\ &=2 \end{aligned} $$

The gradient is $2$. This means $y$ increases by $2$ units for every $1$ unit increase in $x$.

Example 2: Find An Equation From A Point And Gradient

Find the equation of a line with gradient $-3$ passing through $(1,5)$.

Use $y-y_1=m(x-x_1)$:

$$ \begin{aligned} y-5&=-3(x-1) \\ y-5&=-3x+3 \\ y&=-3x+8 \end{aligned} $$

The equation is

$$ y=-3x+8 $$

Check by substituting $(1,5)$:

$$ 5=-3(1)+8=5 $$

Example 3: Write A Perpendicular Line In General Form

A line $L_1$ has equation $2x-3y-4=0$. Find the equation of a line $L_2$ perpendicular to $L_1$ and passing through $(4,-2)$.

First find the gradient of $L_1$:

$$ \begin{aligned} 2x-3y-4&=0 \\ -3y&=-2x+4 \\ y&=\frac{2}{3}x-\frac{4}{3} \end{aligned} $$

So $m_1=\frac{2}{3}$. A perpendicular line has gradient

$$ m_2=-\frac{3}{2} $$

Use the point $(4,-2)$:

$$ \begin{aligned} y-(-2)&=-\frac{3}{2}(x-4) \\ y+2&=-\frac{3}{2}x+6 \\ y&=-\frac{3}{2}x+4 \end{aligned} $$

Multiply by $2$ and rearrange:

$$ \begin{aligned} 2y&=-3x+8 \\ 3x+2y-8&=0 \end{aligned} $$

The required equation is

$$ 3x+2y-8=0 $$

Example 4: Interpret A Line In Context

A water tank contains $20$ litres at the start. Water is added at a constant rate of $5$ litres per minute. Let $y$ be the amount of water after $x$ minutes.

The starting amount is the $y$-intercept, so $c=20$. The rate of increase is the gradient, so $m=5$.

$$ y=5x+20 $$

After $6$ minutes:

$$ \begin{aligned} y&=5(6)+20 \\ &=30+20 \\ &=50 \end{aligned} $$

There will be $50$ litres of water after $6$ minutes.

Example 5: Find An Equation From Two Points

Find the equation of the line through $(2,1)$ and $(6,9)$.

First find the gradient:

$$ \begin{aligned} m&=\frac{9-1}{6-2} \\ &=\frac{8}{4} \\ &=2 \end{aligned} $$

Use the point $(2,1)$ in point-gradient form:

$$ \begin{aligned} y-1&=2(x-2) \\ y-1&=2x-4 \\ y&=2x-3 \end{aligned} $$

Check with the second point:

$$ 9=2(6)-3=9 $$

So the equation is $y=2x-3$.

Example 6: Rearrange General Form And Read Features

For the line $3x+2y-8=0$, find the gradient and $y$-intercept.

Make $y$ the subject:

$$ \begin{aligned} 3x+2y-8&=0 \\ 2y&=-3x+8 \\ y&=-\frac{3}{2}x+4 \end{aligned} $$

So the gradient is $-\frac{3}{2}$ and the $y$-intercept is $4$. The line crosses the $y$-axis at $(0,4)$.

Example 7: Identify A Vertical Line

Find the equation of the line through $(5,-2)$ and $(5,6)$.

Both points have the same $x$ coordinate. The horizontal change is:

$$ 5-5=0 $$

So the usual gradient formula would divide by $0$, which is undefined. The line is vertical, and every point on it has $x=5$. Therefore the equation is:

$$ x=5 $$

Common Mistakes

• Reversing the gradient fraction. Use $\frac{\text{change in }y}{\text{change in }x}$, not $\frac{\text{change in }x}{\text{change in }y}$.
• Mixing coordinates from different points, such as using $y_2-y_1$ with $x_1-x_2$.
• Treating a vertical line as having gradient $0$. A horizontal line has gradient $0$; a vertical line has undefined gradient.
• Forgetting that a negative gradient means the line falls from left to right.
• Reading $c$ in $y=mx+c$ as the $x$-intercept. It is the $y$-intercept.
• Using the same gradient for perpendicular lines. Perpendicular gradients are negative reciprocals when both lines are non-vertical.
• Leaving a general-form answer with fractions when the question expects integer coefficients; multiply through carefully.

Practice Tasks

Foundation

1. Plot the points $(0,1)$, $(2,5)$, and $(4,9)$. Do they appear to lie on one straight line? Explain using gradients.
2. State whether each gradient is positive, negative, zero, or undefined: a line rising left to right; a horizontal line; a vertical line; a line falling left to right.
3. Find the gradient of the line through $(3,7)$ and $(8,17)$.
4. Find the gradient of the line through $(-2,5)$ and $(4,-7)$.

Equations Of Lines

1. Find the gradient and $y$-intercept of $y=-4x+6$.
2. Write the equation of a line with gradient $3$ passing through $(2,-1)$.
3. Find the equation of the line passing through $(1,2)$ and $(5,10)$.
4. Find the equation of the line passing through $(-1,4)$ and $(3,-4)$.
5. Rearrange $4x+2y-10=0$ into the form $y=mx+c$, then state the gradient.
6. Find the $x$-intercept and $y$-intercept of $2x+5y=20$.

Connections And Applications

1. Find the equation of a line parallel to $y=\frac{1}{2}x-7$ passing through $(6,1)$.
2. Find the equation of a line perpendicular to $3x+y-5=0$ passing through $(0,4)$.
3. A taxi fare is $3000$ shillings plus $800$ shillings per kilometre. Write a straight-line equation for the fare $y$ after $x$ kilometres, then find the fare for $12$ kilometres.
4. A water tank starts with $15$ litres and gains $4$ litres each minute. Write the equation for the amount $y$ after $x$ minutes and state the gradient and intercept.
5. A line passes through $(2,3)$ and $(2,9)$. Explain why the usual gradient formula shows that this line is vertical, then write its equation.
6. A learner says the line $y=-5x+2$ has $y$-intercept $-5$. Identify the error and correct it.

Generated Question Layer

• Direct recall: identify gradients, intercepts, coordinate pairs, and line forms.
• Skill fluency: calculate gradients from two points and rearrange equations between $y=mx+c$ and $ax+by+c=0$.
• Graph interpretation: read a gradient and intercept from a drawn straight line.
• Construction: find a line equation from a point and gradient, or from two points.
• Reasoning: compare gradients to decide whether lines are parallel or perpendicular.
• Application: model constant-rate situations such as fares, savings, distance, and liquid levels.
• Error analysis: diagnose a wrong gradient sign, reversed coordinate subtraction, or incorrect perpendicular gradient.

Learner Aid Opportunities

• diagram: labelled Cartesian plane showing axes, quadrants, and sample points.
• graph: dynamic straight line where changing $m$ changes steepness and direction while changing $c$ moves the intercept.
• chart: table linking $y=mx+c$, $y-y_1=m(x-x_1)$, and $ax+by+c=0$ to what each form reveals and when to use it.
• animation: step-by-step rise-over-run movement between two plotted points, with numerator and denominator highlighted.
• interactive: learner enters two points and receives gradient, line equation, intercepts, and a plotted graph.
• interactive: parallel/perpendicular checker where learners choose the matching gradient before writing the line equation.
• video: short worked lesson on converting $ax+by+c=0$ into $y=mx+c$ and reading gradient and intercept.
• LLM tutor: prompt learners to explain why vertical lines have undefined gradient, then challenge reversed-gradient and wrong-intercept misconceptions.

Exam-Derived Signals

• Unreviewed extraction signal: data/topic_frequency_2021_2025.json counts this topic in 2 mapped question records from 2021-2025 Basic Mathematics Paper 1.
• Unreviewed mapped records appear in 2021 and 2022, both involving a line perpendicular to a given straight line and passing through a given point.
• Unreviewed mapping details: records csee_041_2021_p1_q04_a and csee_041_2022_p1_q04_b are marked mapped_unreviewed with needs_manual_review.
• Official exam-format crosswalk signal: the 2022 format group Co-ordinate geometry/[[vector|Vectors]] maps to this topic together with midpoint/distance/parallel/perpendicular lines and vectors. This is assessment guidance, not a replacement for the syllabus.
• Review caution: these signals were read from extracted JSON and have not been checked against the original exam PDFs or marking schemes.

Source And Review Notes

• Curriculum authority: the 2023 CSEE Mathematics syllabus reference at raw/syllabuses/csee/2023/csee_mathematics_syllabus_2023.pdf.
• Registry source: data/curriculum_map.json identifies this as a Form I topic under the coordinate-geometry hub with official review status.
• Assessment enrichment: data/topic_frequency_2021_2025.json, data/question_map_2021_2025.jsonl, and data/exam_format_topic_crosswalk_2022.jsonl provide unreviewed assessment signals only.
• Learner prose, examples, and practice tasks on this page are original draft material for review.
• Review risks: examples should be checked for local notation preferences, expected answer forms, and consistency with teacher-facing schemes before being treated as final.