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Coordinate geometry

IGCSE Mathematics · Topic 3

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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 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

$$y = mx + c,$$

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 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

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 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.

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

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 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:

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

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 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 $-1$:

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

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 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.

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