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Coordinate geometry

IGCSE Mathematics · Topic 3

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Coordinate geometry

Look down on a city like this and every street corner has an address: so many blocks across, so many blocks up. Two numbers, and you know exactly where you…

English narration · English + 中文 subtitles burned in

This handout covers Topic 3, Coordinate geometry. Parts marked (Extended) are only tested on the Extended papers; everything else is for both levels.

3.1

Coordinates

Syllabus
Subject content Notes and examples
Use and interpret Cartesian coordinates in two dimensions.

Source: Cambridge International syllabus

An aerial view of a city street grid
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.

A coordinate grid with labelled x and y axes, the origin O, the four quadrants marked I to IV, and the point (3, -2) plotted
The axes meet at the origin $O$ and split the plane into four quadrants; $(3,-2)$ means 3 right then 2 down.
Explore

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.

Vocabulary Train
English
coordinates/kəʊˈɔːdɪnəts/
axes/ˈæksɪz/
horizontal/ˌhɒrɪˈzɒntl/
vertical/ˈvɜːtɪkl/
origin/ˈɒrɪdʒɪn/
quadrants/ˈkwɒdrənts/
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

$$y = mx + c,$$

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 $5x + 4y = 8$.

$$4y = -5x + 8 \;\Rightarrow\; y = -\tfrac{5}{4}x + 2.$$

So the gradient is $-\tfrac{5}{4}$ and the $y$-intercept is $2$.

Explore

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.

Vocabulary Train
English
gradient/ˈɡreɪdɪənt/
intercept/ˌɪntəˈsept/
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

y = mx + c: gradient and intercept, live
A steep hairpin mountain road
A steep mountain road: gradient measures steepness as rise over run.

The gradient measures how steep a line is:

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

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 $(4, 11)$.

$$m = \frac{11 - 2}{4 - 1} = \frac{9}{3} = 3.$$
A straight line on a grid with the y-intercept marked at c=1 and a gradient triangle showing run 1 and rise 2
For $y=mx+c$, the line crosses the $y$-axis at $c$ and the gradient $m$ is the rise divided by the run.
Explore

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:

  1. Mark the intercept $c$ on the $y$-axis.
  2. Use the gradient to step to more points (for $m = 3$, go $1$ right and $3$ up).
  3. 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 y equals 3x minus 1: mark the intercept at minus 1, step 1 right and 3 up twice to get more points, then join them with a line
Drawing a line fast: mark the intercept, then use the gradient to step to new points
Vocabulary Train
English
table of values/ˈteɪbl ɒv ˈvæljuːz/
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.

$$y = 3x + c, \qquad 2 = 3(1) + c, \qquad c = -1.$$

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:

$$\text{length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$

To find the midpoint 中点 (the point exactly in the middle), average the coordinates:

$$\text{midpoint} = \left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2} \right).$$

Worked example. Find the length and midpoint of the segment from $(1, 2)$ to $(4, 6)$.

$$\text{length} = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$$
$$\text{midpoint} = \left( \frac{1 + 4}{2}, \; \frac{2 + 6}{2} \right) = (2.5, \, 4).$$
The segment from (1,2) to (4,6) on a grid, with a right triangle of legs 3 and 4, length 5, and the midpoint (2.5, 4) marked
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.
Explore

Length and midpoint lab

midpoint is halfway between endpoints

Move along a line segment and see midpoint as halfway.

Vocabulary Train
English
line segment/laɪn ˈseɡmənt/
length/leŋθ/
Pythagoras' theorem/paɪˈθæɡərəs ˈθɪərəm/
midpoint/ˈmɪdpɔɪnt/
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:

$$-3 = 4(1) + c \;\Rightarrow\; c = -7,$$

so the line is $y = 4x - 7$.

Three parallel lines with gradient 4 and different intercepts; the line through the marked point 1, minus 3 is y equals 4x minus 7
Parallel lines: same gradient, different intercepts
Explore

Parallel & perpendicular

y = ax + b

Parallel lines share a gradient; perpendicular gradients multiply to −1.

Vocabulary Train
English
parallel/ˈpærəlel/
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 $-1$:

$$m_1 \times m_2 = -1, \qquad \text{so} \qquad m_2 = -\frac{1}{m_1}.$$

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 $-\tfrac{2}{3}$.

Two side-by-side grids: on the left two parallel lines with the same gradient, on the right two perpendicular lines meeting at a right angle
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

$$y = \tfrac{6}{5}x - \tfrac{3}{5}.$$
Vocabulary Train
English
perpendicular/ˌpɜːpənˈdɪkjʊlə/
right angle/raɪt ˈæŋɡl/
perpendicular bisector/ˌpɜːpənˈdɪkjʊlə baɪˈsektə/
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.

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