Straight-Line Graphs & Coordinate Geometry

Coordinates and the number plane

Every point on a graph is fixed by a pair of coordinates $(x, y)$. The first number is the horizontal position (across), the second is the vertical position (up or down). They are measured from the origin $(0, 0)$, where the two axes cross.

The plane is split into four quadrants. Reading signs carefully matters: $(3, -2)$ sits bottom-right, while $(-3, 2)$ sits top-left.

Key terms Origin — the point $(0,0)$ where the axes meet.

x-coordinate — horizontal distance (the first number).

y-coordinate — vertical distance (the second number).

Midpoint of a line segment

The midpoint of the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ is the average of the coordinates:

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

For example, the midpoint of $A(2, 1)$ and $B(6, 9)$ is

$$\left( \frac{2+6}{2}, \frac{1+9}{2} \right) = (4, 5).$$

Exam tip Add the two $x$-values then halve; do the same with the $y$-values. A quick check: the midpoint should lie between the two points, so its coordinates sit between theirs.

Length of a line segment (distance formula)

The distance between two points comes straight from Pythagoras' theorem. The horizontal gap is $(x_2 - x_1)$ and the vertical gap is $(y_2 - y_1)$, so the segment is the hypotenuse of a right-angled triangle:

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

For $A(2, 1)$ and $B(6, 9)$:

$$d = \sqrt{(6-2)^2 + (9-1)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}.$$

Watch out The differences are squared, so a negative gap causes no trouble: $(-4)^2 = 16$. But never simplify $\sqrt{16+64}$ to $\sqrt{16}+\sqrt{64}$ — you must add inside the root first.

Gradient of a line

The gradient $m$ measures steepness: how far the line rises for each step across.

$$m = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$$

A positive gradient rises from left to right (uphill).
A negative gradient falls from left to right (downhill).
A horizontal line has gradient $0$; a vertical line has an undefined gradient.

The equation $y = mx + c$

Every straight line can be written as

$$y = mx + c$$

where $m$ is the gradient and $c$ is the y-intercept — the $y$-value where the line crosses the $y$-axis (at $x = 0$).

So $y = 2x - 3$ has gradient $2$ and crosses the $y$-axis at $(0, -3)$. To read these off, the equation must be rearranged into this form first.

Worked example Find the gradient and $y$-intercept of $3x + 2y = 12$.

Make $y$ the subject:

$2y = -3x + 12$

Finding the equation of a line through points

Given the gradient and one point: substitute into $y = mx + c$ and solve for $c$.

Find the line with gradient $4$ through $(2, 5)$:

$$5 = 4(2) + c \;\Rightarrow\; 5 = 8 + c \;\Rightarrow\; c = -3$$

So the equation is $y = 4x - 3$.

Given two points: first find the gradient, then use one point as above.

Worked example Find the equation of the line through $A(1, 2)$ and $B(4, 11)$.

Gradient:

$m = \dfrac{11 - 2}{4 - 1} = \dfrac{9}{3} = 3$

Parallel and perpendicular lines

Parallel lines never meet, so they have the same gradient. Lines $y = 2x + 1$ and $y = 2x - 5$ are parallel because both have $m = 2$.

Perpendicular lines cross at $90°$. Their gradients multiply to $-1$:

$$m_1 \times m_2 = -1$$

So the perpendicular gradient is the negative reciprocal: flip the fraction and change the sign. If $m_1 = 3$, then $m_2 = -\tfrac{1}{3}$. If $m_1 = -\tfrac{2}{5}$, then $m_2 = \tfrac{5}{2}$.

Worked example Find the line perpendicular to $y = \tfrac{1}{2}x + 4$ passing through $(3, 1)$.

Gradient of the given line is $\tfrac{1}{2}$, so the perpendicular gradient is the negative reciprocal, $-2$.

Drawing straight-line graphs

To plot a line from its equation, build a small table of values. Choose a few $x$-values, work out each $y$, plot the points and join them with a straight ruled line.

For $y = 2x - 1$:

A faster method when the equation is in $y = mx + c$ form: mark the intercept $c$ on the $y$-axis, then use the gradient (rise over run) to step to the next point and draw through both.

Exam tip Lines like $x = 4$ are vertical and $y = -2$ are horizontal — students often swap these. Remember $x = 4$ means "$x$ is always $4$", a vertical line through $(4, 0)$.

Solving simultaneous equations graphically

Two straight lines drawn on the same axes meet at one point (unless they are parallel). The coordinates of that intersection are the solution to the pair of simultaneous equations, because that point lies on both lines at once.

To solve graphically:

1. Draw both lines accurately on one set of axes.
2. Read off the coordinates of the crossing point.
3. State the solution as $x = \dots$, $y = \dots$.

For example, $y = x + 1$ and $y = -x + 5$ cross at $(2, 3)$, so the solution is $x = 2$, $y = 3$. You can confirm this by checking both equations: $3 = 2 + 1$ ✓ and $3 = -2 + 5$ ✓.

Real world Intersection points model break-even analysis. If one line is a company's costs and another its income against units sold, the crossing point is the break-even quantity — sell more than that and you make a profit.

Watch out If two lines have the same gradient they are parallel and never cross, so the simultaneous equations have no solution. Identical equations give the same line and infinitely many solutions.