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E3.1 Coordinates

A Cartesian coordinate $(x,y)$ records a point’s horizontal and vertical positions in that order. Accurate interpretation depends on order, sign and the scale of each axis.

Coordinate	Positive direction	Negative direction
$x$	right	left
$y$	up	down

To read a point, project to the $x$ -axis and then the $y$ -axis. To plot it, start at $(0,0)$ , move horizontally by $x$ , then vertically by $y$ . Read each axis scale independently before counting grid spaces.

Location	Coordinate condition
on the $x$ -axis	$y=0$
on the $y$ -axis	$x=0$
at the origin	$(0,0)$

Coordinates can also encode a shape. Compare corresponding coordinate changes: moving from $A(x_1,y_1)$ to $B(x_2,y_2)$ changes $x$ by $x_2-x_1$ and $y$ by $y_2-y_1$ . Apply the same changes to another vertex when equal, parallel sides require the same displacement.

$A(1,1),\ B(4,2),\ D(2,5):\quad C=B+(D-A)=(4,2)+(1,4)=(5,6)$

Check any inferred point against every stated condition: coordinate signs, whole-number restrictions, equal side movement, and the order in which vertices are named.

Do not swap coordinates or treat a shape sketch as accurately scaled. Derive positions from the grid values and stated geometric relationships.

E3.2 Drawing linear graphs

A linear equation describes every point $(x,y)$ on one straight line. To draw the graph, find coordinate pairs that satisfy the equation, plot them accurately and join them with a ruled line.

Choose several convenient $x$ -values within the grid range.
Calculate the matching $y$ -value for each one. Rearrange first if necessary.
Plot the resulting $(x,y)$ points, using each axis scale correctly.
Check that the points line up, then draw one straight ruled line through them.

$3x+2y=5\quad\Longrightarrow\quad y=\frac{5-3x}{2}$

$x$	$-1$	$1$	$3$
$y=(5-3x)/2$	$4$	$1$	$-2$
point	$(-1,4)$	$(1,1)$	$(3,-2)$

For an equation such as $x+y=7$ , setting $x=0$ gives $(0,7)$ and setting $y=0$ gives $(7,0)$ . These two intercepts determine the line; an extra point is a useful check.

Substitute a plotted point back into the original equation. For $(3,-2)$ , $3(3)+2(-2)=5$ , so the point belongs on $3x+2y=5$ .

Two distinct correct points determine a straight line, but a third point helps reveal an arithmetic or plotting error. Do not draw separate segments, use a freehand curve, or confuse $x=7$ or $y=7$ with $x+y=7$ .

E3.3 Gradient of linear graphs

The gradient measures how much $y$ changes for each unit increase in $x$ . Read the line from left to right: a rising line has positive gradient and a falling line has negative gradient.

$\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}=\frac{\text{rise}}{\text{run}}$

Choose two clear grid points on the line. From the left point to the right point, record the horizontal change first and the signed vertical change second. Divide vertical change by horizontal change using the actual axis values, not merely the number of grid squares.

If moving $4$ units right along a line requires moving $3$ units down, the horizontal change is $+4$ and the vertical change is $-3$ . The gradient is therefore $-3/4=-0.75$ .

A horizontal line has gradient $0$ because its vertical change is $0$ . A vertical line has no defined gradient because its horizontal change is $0$ , which would require division by zero.

Do not reverse the fraction or discard the sign. Any two exact points on the same straight line give the same gradient, but estimated or misread points can change the result.

For two points on a straight line, calculate gradient by dividing their change in $y$ by their change in $x$ . Both differences must use the same point order.

$m=\frac{y_2-y_1}{x_2-x_1}\qquad\text{for }(x_1,y_1)\text{ and }(x_2,y_2)$

Label the two coordinates, subtract the $y$ -values for the numerator, then subtract the corresponding $x$ -values in the same order for the denominator. Keep brackets around negative coordinates before simplifying.

$(-2,-7),\ (4,1):\qquad m=\frac{1-(-7)}{4-(-2)}=\frac{8}{6}=\frac{4}{3}$

Reversing both subtractions gives $(-7-1)/(-2-4)=(-8)/(-6)=4/3$ , so the gradient is unchanged. Reversing only one subtraction incorrectly changes the sign.

If the two $x$ -coordinates are equal, the denominator is zero and the line is vertical, so its gradient is undefined. Otherwise give the fraction in its simplest form unless a decimal is required.

E3.4 Length and midpoint

The horizontal and vertical changes between two points form the perpendicular sides of a right-angled triangle. The line segment is its hypotenuse, so its length follows from Pythagoras’ theorem.

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

Subtract corresponding coordinates to find $\Delta x$ and $\Delta y$ . Keep brackets around negative values, square both changes, add, then take the positive square root. The subtraction order does not affect the length because each difference is squared.

$A(-5,2),\ B(7,8):\quad d=\sqrt{(7-(-5))^2+(8-2)^2}=\sqrt{180}=6\sqrt5\approx13.4$

The answer must be at least as large as the greater of $|\Delta x|$ and $|\Delta y|$ . Here $13.4>12$ , so the result is plausible. If the points share an $x$ - or $y$ -coordinate, the formula reduces to the absolute difference on the other axis.

Do not add coordinate differences directly or round intermediate values. Square each complete signed difference and round only the final decimal when the required accuracy is known.

The midpoint lies exactly halfway between the endpoints in both horizontal and vertical directions. Find it by averaging the two $x$ -coordinates and averaging the two $y$ -coordinates separately.

$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Pair like coordinates: add $x$ to $x$ and divide by $2$ , then add $y$ to $y$ and divide by $2$ . Use brackets when a coordinate is negative so the sign remains part of the sum.

$P(-1,3),\ Q(6,4):\quad M=\left(\frac{-1+6}{2},\frac{3+4}{2}\right)=(2.5,3.5)$

From $P$ to $M$ , the change is $(3.5,0.5)$ ; from $M$ to $Q$ , it is also $(3.5,0.5)$ . Equal coordinate changes confirm that $M$ is halfway along the segment.

Do not divide only one coordinate or average an $x$ -coordinate with a $y$ -coordinate. A midpoint can have halves or other decimals even when both endpoints have integer coordinates.

E3.5 Equations of linear graphs

A straight-line equation links every point on the line. In $y=mx+c$ , $m$ is the gradient and $c$ is the $y$ -coordinate where the line crosses the $y$ -axis, so the intercept point is $(0,c)$ .

Form	How to interpret it
$y=mx+c$	read gradient $m$ and $y$ -intercept $(0,c)$ directly
$ax+by=c$ , $b\ne0$	rearrange to make $y$ the subject
$x=k$	vertical line through every point with $x$ -coordinate $k$ ; gradient undefined

$5x+4y=8\quad\Longrightarrow\quad y=-\frac54x+2\quad\Longrightarrow\quad m=-\frac54,\ (0,2)$

To obtain an equation from a graph or two points: find the gradient $m$ ; substitute one known point into $y=mx+c$ to calculate $c$ ; then write and fully simplify the equation. A graph may allow $c$ to be read directly from the $y$ -axis.

$A(3,16),\ B(8,31):\quad m=\frac{31-16}{8-3}=3,\quad 16=3(3)+c\Rightarrow c=7,\quad y=3x+7$

Substitute the other point: $3(8)+7=31$ , so both $A$ and $B$ satisfy the equation. When a requested form is specified, rearrange the final result into that form and remove common factors or unnecessary signs.

Do not confuse the intercept value $c$ with the point $(0,c)$ , and do not force a vertical line into $y=mx+c$ . Also distinguish $y=5-3x$ from $y=5x-3$ : coefficients and constants have different roles.

E3.6 Parallel lines

Distinct non-vertical parallel lines have the same gradient because they rise or fall at the same rate, but they have different intercepts because they occupy different positions.

$y=m_1x+c_1\ \parallel\ y=m_2x+c_2\quad\Longrightarrow\quad m_1=m_2$

Rearrange the given line into $y=mx+c$ if needed and identify $m$ .
Write the parallel line as $y=mx+k$ with the same gradient.
Substitute the point on the new line to calculate $k$ .
Simplify the equation and check both the gradient and the given point.

$y=4x-1,\ (1,-3):\quad -3=4(1)+k\Rightarrow k=-7\Rightarrow y=4x-7$

If a parallel line has gradient $1/7$ and crosses the $x$ -axis at $x=2$ , it passes through $(2,0)$ . Substitution gives $0=(1/7)(2)+k$ , so $k=-2/7$ and $y=(1/7)x-2/7$ , equivalently $7y=x-2$ .

The new line has the required gradient and its stated point satisfies the equation. In the syllabus example, substituting $(1,-3)$ into $y=4x-7$ gives $-3=4-7$ .

Do not copy the original intercept: that would reproduce the same line, not a distinct parallel one. Vertical lines are written $x=k$ ; two distinct vertical lines are parallel even though their gradients are undefined.

E3.7 Perpendicular lines

Two non-vertical, non-horizontal perpendicular lines meet at a right angle and have gradients that are negative reciprocals. This reverses the rise/run ratio and changes its sign.

$m_1m_2=-1\qquad\Longleftrightarrow\qquad m_2=-\frac{1}{m_1}$

First expose the given gradient. From $2y=3x+1$ , obtain $y=(3/2)x+1/2$ , so $m_1=3/2$ . The perpendicular gradient is $m_2=-2/3$ .

For a perpendicular line through a specified point, write $y=m_2x+c$ , substitute the point to find $c$ , then simplify and check. For example, perpendicular to $y=- frac12x+7$ through $(3,5)$ gives $m_2=2$ and $5=2(3)+c$ , so $y=2x-1$ .

A perpendicular bisector must be both perpendicular to the segment and pass through its midpoint. Find the segment gradient, take its negative reciprocal, calculate the midpoint, then substitute that midpoint into the new line equation.

$A(-3,8),\ B(9,-2):\quad M=(3,3),\quad m_{AB}=-\frac56,\quad m_{\perp}=\frac65,\quad y=\frac65x-\frac35$

The final line has gradient $6/5$ and contains $(3,3)$ because $(6/5)(3)-3/5=3$ . Its gradient product with $-5/6$ is $-1$ , confirming perpendicularity.

Do not only change the sign: the reciprocal is also required. Horizontal and vertical lines are the special pair— $y=k$ is perpendicular to $x=a$ —so the product rule is not used when a gradient is undefined.

3. Coordinate geometry

Exam analysis

Most tested topics

E3.1 Coordinates

Use coordinates to locate and relate points

E3.2 Drawing linear graphs

Draw a straight line from its equation

E3.3 Gradient of linear graphs

Read a line’s gradient as rise over run

Calculate gradient from two coordinates

E3.4 Length and midpoint

Calculate the length between two points

Find the midpoint of a line segment

E3.5 Equations of linear graphs

Interpret and find a straight-line equation

E3.6 Parallel lines

Find the equation of a parallel line

E3.7 Perpendicular lines

Find perpendicular lines and bisectors