This handout covers Topic 3, Coordinate geometry. Parts marked (Extended) are only tested on the Extended papers; everything else is for both levels.
Coordinate geometry
IGCSE Mathematics · Topic 3
3.1
Coordinates
Syllabus
| Subject content | Notes and examples |
|---|---|
| Use and interpret Cartesian coordinates in two dimensions. |
Source: Cambridge International syllabus
A city street grid: every place is fixed by its coordinates.
A point on a graph is described by its coordinates 坐标, sometimes called Cartesian coordinates, written $(x, y)$. The first number is the across value and the second is the up value.
- The two number lines are the axes 坐标轴: the horizontal 水平 $x$-axis and the vertical 竖直 $y$-axis.
- They cross at the origin 原点, the point $(0, 0)$.
- The axes split the grid into four quadrants 象限.
So the point $(3, -2)$ is found by going $3$ to the right and $2$ down.
The axes meet at the origin $O$ and split the plane into four quadrants; $(3,-2)$ means 3 right then 2 down.
The coordinate plane
y = mx + c
Every point has an (x, y) coordinate. A straight line is the set of points where y depends on x in a fixed way.
| English | Chinese | Pinyin |
|---|---|---|
| coordinates | 坐标 | zuò biāo |
| axes | 坐标轴 | zuò biāo zhóu |
| horizontal | 水平 | shuǐ píng |
| vertical | 竖直 | shù zhí |
| origin | 原点 | yuán diǎn |
| quadrants | 象限 | xiàng xiàn |
3.5
The equation of a straight line
Syllabus
| Subject content | Notes and examples |
|---|---|
| Interpret and obtain the equation of a straight-line graph in the form $y = mx + c$. | Questions may: • use and request lines in the forms $y = mx + c$, $x = k$ • involve finding the equation when the graph is given • ask for the gradient or $y$-intercept of a graph from an equation, e.g. find the gradient and $y$-intercept of the graph with the equation $y = 6x + 3$. Candidates are expected to give equations of a line in a fully simplified form. |
| Subject content | Notes and examples |
|---|---|
| Interpret and obtain the equation of a straight-line graph. | Questions may: • use and request lines in different forms, e.g. $ax + by = c$, $y = mx + c$, $x = k$ • involve finding the equation when the graph is given • ask for the gradient or $y$-intercept of a graph from an equation, e.g. find the gradient and $y$-intercept of the graph with equation $5x + 4y = 8$. Candidates are expected to give equations of a line in a fully simplified form. |
Source: Cambridge International syllabus
Most straight lines can be written as
where $m$ is the gradient 斜率 (steepness) and $c$ is the intercept 截距 — the $y$-value where the line crosses the $y$-axis.
- A line like $x = k$ (for example $x = 3$) is vertical.
- A line like $y = k$ (for example $y = 3$) is horizontal.
A line may also be given as $ax + by = c$. Rearrange it into $y = mx + c$ to read off the gradient and intercept.
Worked example. Find the gradient and $y$-intercept of $5x + 4y = 8$.
So the gradient is $-\tfrac{5}{4}$ and the $y$-intercept is $2$.
y = mx + c
y = ax + b
Drag the gradient and the intercept. a is the gradient (steepness) and b is where the line crosses the y-axis.
| English | Chinese | Pinyin |
|---|---|---|
| gradient | 斜率 | xié lǜ |
| intercept | 截距 | jié jù |
3.3
Gradient
Syllabus
| Subject content | Notes and examples |
|---|---|
| Find the gradient of a straight line. | From a grid only. |
| Subject content | Notes and examples |
|---|---|
| 1 Find the gradient of a straight line. | |
| 2 Calculate the gradient of a straight line from the coordinates of two points on it. |
Source: Cambridge International syllabus
A steep mountain road: gradient measures steepness as rise over run.
The gradient measures how steep a line is:
A positive gradient goes up to the right; a negative gradient goes down to the right.
Worked example. Find the gradient of the line through $(1, 2)$ and $(4, 11)$.
For $y=mx+c$, the line crosses the $y$-axis at $c$ and the gradient $m$ is the rise divided by the run.
Gradient
y = ax + b
The gradient a measures steepness — rise over run.
3.2
Drawing a straight-line graph
Syllabus
| Subject content | Notes and examples |
|---|---|
| Draw straight-line graphs for linear equations. | Equations will be given in the form $y = mx + c$ (e.g. $y = -2x + 5$), unless a table of values is given. |
| Subject content | Notes and examples |
|---|---|
| Draw straight-line graphs for linear equations. | Examples include: • $y = -2x + 5$ • $y = 7 - 4x$ • $3x + 2y = 5$. |
Source: Cambridge International syllabus
To draw $y = mx + c$, the quickest way is:
- Mark the intercept $c$ on the $y$-axis.
- Use the gradient to step to more points (for $m = 3$, go $1$ right and $3$ up).
- Join the points with a straight line.
You can also make a small table of values 数值表: choose two or three $x$-values, work out $y$, and plot the points.
Drawing a line fast: mark the intercept, then use the gradient to step to new points
| English | Chinese | Pinyin |
|---|---|---|
| table of values | 数值表 | shù zhí biǎo |
3.5
Finding the equation of a line
Syllabus
| Subject content | Notes and examples |
|---|---|
| Interpret and obtain the equation of a straight-line graph in the form $y = mx + c$. | Questions may: • use and request lines in the forms $y = mx + c$, $x = k$ • involve finding the equation when the graph is given • ask for the gradient or $y$-intercept of a graph from an equation, e.g. find the gradient and $y$-intercept of the graph with the equation $y = 6x + 3$. Candidates are expected to give equations of a line in a fully simplified form. |
| Subject content | Notes and examples |
|---|---|
| Interpret and obtain the equation of a straight-line graph. | Questions may: • use and request lines in different forms, e.g. $ax + by = c$, $y = mx + c$, $x = k$ • involve finding the equation when the graph is given • ask for the gradient or $y$-intercept of a graph from an equation, e.g. find the gradient and $y$-intercept of the graph with equation $5x + 4y = 8$. Candidates are expected to give equations of a line in a fully simplified form. |
Source: Cambridge International syllabus
If you know the gradient $m$ and one point on the line, put the point into $y = mx + c$ to find $c$.
Worked example. A line has gradient $3$ and passes through $(1, 2)$. Find its equation.
So the equation is $y = 3x - 1$. (If you are given two points, first find the gradient, then do this.)
3.4
Length and midpoint (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Calculate the length of a line segment. | |
| 2 Find the coordinates of the midpoint of a line segment. |
Source: Cambridge International syllabus
A line segment 线段 is the straight piece between two points.
To find the length 长度 of the segment between $(x_1, y_1)$ and $(x_2, y_2)$, use Pythagoras' theorem 勾股定理 on the horizontal and vertical gaps:
To find the midpoint 中点 (the point exactly in the middle), average the coordinates:
Worked example. Find the length and midpoint of the segment from $(1, 2)$ to $(4, 6)$.
The horizontal gap (3) and vertical gap (4) make a right triangle, so the length is $\sqrt{3^2+4^2}=5$; the midpoint is the average of the coordinates.
Length and midpoint lab
midpoint is halfway between endpoints
Move along a line segment and see midpoint as halfway.
| English | Chinese | Pinyin |
|---|---|---|
| line segment | 线段 | xiàn duàn |
| length | 长度 | cháng dù |
| Pythagoras' theorem | 勾股定理 | gōu gǔ dìng lǐ |
| midpoint | 中点 | zhōng diǎn |
3.6
Parallel lines
Syllabus
| Subject content | Notes and examples |
|---|---|
| Find the gradient and equation of a straight line parallel to a given line. | e.g. find the equation of the line parallel to $y = 4x - 1$ that passes through $(1, -3)$. |
Source: Cambridge International syllabus
Parallel 平行 lines never meet, so they have the same gradient.
Worked example. Find the equation of the line parallel to $y = 4x - 1$ that passes through $(1, -3)$.
The gradient is also $4$. Put the point in:
so the line is $y = 4x - 7$.
Parallel lines: same gradient, different intercepts
Parallel & perpendicular
y = ax + b
Parallel lines share a gradient; perpendicular gradients multiply to −1.
| English | Chinese | Pinyin |
|---|---|---|
| parallel | 平行 | píng xíng |
3.7
Perpendicular lines (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| Find the gradient and equation of a straight line perpendicular to a given line. | Examples include: • find the gradient of a line perpendicular to $2y = 3x + 1$ • find the equation of the perpendicular bisector of the line joining the points $(-3, 8)$ and $(9, -2)$. |
Source: Cambridge International syllabus
Two lines are perpendicular 垂直 if they meet at a right angle 直角. Their gradients multiply to $-1$:
In words: flip the fraction and change the sign.
Worked example. Find the gradient of a line perpendicular to $2y = 3x + 1$.
Rearrange: $y = \tfrac{3}{2}x + \tfrac{1}{2}$, so the gradient is $\tfrac{3}{2}$. The perpendicular gradient is $-\tfrac{2}{3}$.
Parallel lines share the same gradient; perpendicular gradients multiply to $-1$.
Perpendicular bisector
The perpendicular bisector 垂直平分线 of a segment cuts it in half at a right angle. To find its equation: get the midpoint, then use the perpendicular gradient through that midpoint.
Worked example. Find the perpendicular bisector of the segment joining $(-3, 8)$ and $(9, -2)$.
- Midpoint: $\left( \frac{-3 + 9}{2}, \frac{8 + (-2)}{2} \right) = (3, 3)$.
- Gradient of the segment: $\frac{-2 - 8}{9 - (-3)} = \frac{-10}{12} = -\tfrac{5}{6}$.
- Perpendicular gradient: $\frac{6}{5}$.
Through $(3, 3)$: $\; 3 = \tfrac{6}{5}(3) + c \Rightarrow c = 3 - \tfrac{18}{5} = -\tfrac{3}{5}$. So the bisector is
| English | Chinese | Pinyin |
|---|---|---|
| perpendicular | 垂直 | chuí zhí |
| right angle | 直角 | zhí jiǎo |
| perpendicular bisector | 垂直平分线 | chuí zhí píng fēn xiàn |
3.7
Exam tips
- The straight line is $y = mx + c$: $m$ is the gradient and $c$ is where the line crosses the $y$-axis.
- Gradient = (change in $y$) ÷ (change in $x$). Keep the two coordinates in the same order on the top and the bottom.
- Parallel lines have the same gradient; perpendicular lines have gradients that multiply to $-1$ (the negative reciprocal).
- The midpoint is the average of the coordinates; the distance between two points comes from Pythagoras on the differences.