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Straight-Line Graphs

Q: How do I find the equation of a line from two points?

First find the gradient with m = y_2-y_1x_2-x_1, then substitute the gradient and either point into y - y_1 = m(x - x_1) and rearrange to the form asked for.

Q: When should I use y = mx + c versus y - y_1 = m(x - x_1)?

Use y = mx + c when you know the y-intercept directly. Use the point-gradient form when you know the gradient and a general point — it avoids solving for c separately.

A straight line is captured entirely by two numbers: its gradient and where it sits. From those you can write its equation, test whether two lines are parallel or perpendicular, and find lengths and midpoints.

28 min Video by TLMaths Coordinate Geometry

The big picture

Coordinate geometry is where algebra and geometry meet: shapes become equations, and geometric questions (“are these lines perpendicular?”, “where do they cross?”) become algebra you can solve. The straight line is the foundation of all of it. The gradient idea you meet here — rate of change of $y$ with $x$ — is exactly what differentiation generalises to curves, so time spent here pays off directly in calculus. And the parallel/perpendicular conditions reappear whenever you find a tangent or a normal to a curve.

A-Level Maths C1-07: The Equation of a Line in the form y = mx + cWatch the walkthrough, then work through the notes below.

Open on YouTube

What you'll be able to do

Find the gradient between two points and interpret it as a rate
Write the equation of a line in $y = m x + c$ and $y - y_{1} = m (x - x_{1})$ forms
Use the parallel ( $m_{1} = m_{2}$ ) and perpendicular ( $m_{1} m_{2} = - 1$ ) conditions
Find the midpoint and length of a line segment

Gradient: steepness as a number

The $gradient$ measures how steeply a line climbs: the change in $y$ divided by the change in $x$ between any two points on it — “rise over run”. A positive gradient rises left-to-right, a negative one falls, and zero is flat.

Because a straight line has the $same$ gradient everywhere, you can compute it from any two points and it will always agree. That constancy is precisely what makes the line straight.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Change in

y

over change in

x

— keep the points in the same order top and bottom.

1Rise:

11 - 2 = 9

. Run:

4 - 1 = 3

m = \frac{9}{3} = 3

Answer

m = 3

Why this works

Gradient is a rate: “for every 1 across, 3 up”. This is the same idea as speed (distance per time) or the derivative later — a straight line is just the case where the rate never changes.

Writing the equation of a line

The familiar form $y = m x + c$ names the gradient $m$ and the $y$ -intercept $c$ . But when you know a gradient and a $point$ (not the intercept), the point-gradient form is faster: $y - y_{1} = m (x - x_{1})$ . Substitute and rearrange to whatever form the question wants.

y = m x + c

Gradient

m

y

-intercept

c

y - y_{1} = m (x - x_{1})

Best when you know the gradient and one point.

1Use point-gradient:

y - 1 = 2 (x - 3)

2Expand:

y - 1 = 2 x - 6

3Rearrange:

y = 2 x - 5

Answer

y = 2 x - 5

Tip — Exam answers are often wanted as $a x + b y + c = 0$ with integer coefficients — be ready to clear fractions and rearrange.

Parallel and perpendicular lines

Two lines are $parallel$ exactly when they have the $same gradient$ . They are $perpendicular$ when their gradients multiply to $- 1$ — equivalently, each is the $negative reciprocal$ of the other. So a line perpendicular to one of gradient 2 has gradient $- \frac{1}{2}$ .

m_{1} = m_{2} (parallel)

Equal gradients never meet.

m_{1} m_{2} = - 1 (perpendicular)

Negative reciprocal: flip and change sign.

1The gradient is

\frac{3}{4}

2Perpendicular gradient is the negative reciprocal: flip to

\frac{4}{3}

, change sign to

- \frac{4}{3}

Answer

- \frac{4}{3}

Why this works

The $m_{1} m_{2} = - 1$ rule is what lets you find a $normal$ to a curve later: differentiate for the tangent gradient, then take the negative reciprocal for the perpendicular normal.

Midpoint and length

Two more essentials. The $midpoint$ of a segment is the average of the endpoints. The $length$ comes straight from Pythagoras on the horizontal and vertical gaps.

midpoint = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

Average the coordinates.

length = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Pythagoras on the gaps.

1Midpoint:

(\frac{1 + 4}{2}, \frac{2 + 6}{2}) = (2.5, 4)

2Length:

(4 - 1)^{2} + (6 - 2)^{2} = 9 + 16 = 25 = 5

AnswerMidpoint

(2.5, 4)

, length

5

Think like an examiner

For a perpendicular line, use the negative reciprocal — a sign slip or forgetting to flip is the classic lost mark.
If the answer is required as $a x + b y + c = 0$ , clear fractions so the coefficients are integers.
Keep the points in a consistent order in the gradient formula — swapping one but not the other flips the sign.
A “normal” to a curve is a perpendicular line; expect to combine differentiation with the $m_{1} m_{2} = - 1$ rule.

Common misconceptions

Taking the perpendicular gradient of 2 as

- 2

It is the negative reciprocal,

- \frac{1}{2}

: flip and change sign.

Computing gradient as

\frac{x _{2} - x _{1}}{y _{2} - y _{1}}

It is

\frac{y _{2} - y _{1}}{x _{2} - x _{1}}

— rise over run, not run over rise.

Mismatching the order of points, e.g.

y_{2} - y_{1}

over

x_{1} - x_{2}

Subtract in the same order top and bottom, or the sign will be wrong.

Forgetting the square root in the length formula.

Length

= (Δ x)^{2} + (Δ y)^{2}

— the squared distance is not the distance.

Straight-line toolkit

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Gradient.

y - y_{1} = m (x - x_{1})

Point-gradient form.

m_{1} m_{2} = - 1

Perpendicular condition.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Distance between points.

Stretch yourself

The points $A (1, 2)$ , $B (5, 4)$ are given. Find the equation of the perpendicular bisector of $A B$ .

Hint — The perpendicular bisector passes through the midpoint of $A B$ with a gradient perpendicular to $A B$ .

Questions students ask

How do I find the equation of a line from two points?

First find the gradient with

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

, then substitute the gradient and either point into

y - y_{1} = m (x - x_{1})

and rearrange to the form asked for.

What is the gradient of a line perpendicular to another?

The negative reciprocal: flip the fraction and change the sign. So perpendicular to gradient

\frac{3}{4}

- \frac{4}{3}

, because their product is

- 1

When should I use

y = m x + c

versus

y - y_{1} = m (x - x_{1})

Use

y = m x + c

when you know the

y

-intercept directly. Use the point-gradient form when you know the gradient and a general point — it avoids solving for

c

separately.

What is a perpendicular bisector?

The line that cuts a segment in half at right angles. It passes through the midpoint with the negative-reciprocal gradient, and every point on it is equidistant from the two ends.

Key takeaways

Gradient $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ measures steepness as a rate.
Write lines as $y = m x + c$ or $y - y_{1} = m (x - x_{1})$ .
Parallel: equal gradients. Perpendicular: gradients multiply to $- 1$ (negative reciprocal).
Midpoint averages the coordinates; length uses Pythagoras.
These ideas power tangents and normals to curves later.

How this fits the course

Build on

Coordinates and graphs (GCSE)soon

Circles

Leads to

Circles
Differentiation — tangents & normals (later)soon

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