1.03aPureFoundation

Straight Lines

Q: What is the gradient of a vertical line?

It has none. The change in x is zero, so the fraction is undefined. Vertical lines are written x = k and sit outside the gradient formulae entirely.

OCR expects fluency with the equation of a straight line "in the forms $y = m x + c$ and $y - y_{1} = m (x - x_{1})$ ", plus the conditions for lines to be parallel or perpendicular. One quantity ties all of that together: the gradient.

30 min Coordinate Geometry

The big picture

Straight lines are the foundation the rest of coordinate geometry is built on, and they never stop being useful. A tangent to a curve is a straight line, so every differentiation question that asks for a tangent or normal ends up here. Circle questions lean on perpendicular gradients because a tangent meets the radius at right angles. Later, log-linear modelling deliberately transforms an exponential relationship into a straight line so that the gradient and intercept can be read off. Straight-line technique is not an early topic you leave behind — it is the tool you keep returning to with harder inputs.

What you'll be able to do

Calculate a gradient from two points and interpret its sign and size
Find the equation of a line through a point with a known gradient
Apply the parallel and perpendicular gradient conditions
Find midpoints, distances and points of intersection

Gradient: the number that defines the line

The gradient measures how steeply a line rises for each unit it moves right: $m = \frac{change in y}{change in x}$ . A positive gradient rises left to right, a negative one falls, and a larger size means steeper.

Subtract the coordinates in the $same order$ top and bottom. Doing $y_{2} - y_{1}$ over $x_{1} - x_{2}$ produces the right size with the wrong sign, which is the most common slip in the topic.

A horizontal line has gradient 0 and equation $y = k$ . A vertical line has no gradient at all — the denominator would be zero — and its equation is $x = k$ . Vertical lines are the exception every general formula quietly excludes.

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

Gradient between two points, subtracting consistently.

1Take

B

as point 2:

y_{2} - y_{1} = - 3 - 5 = - 8

2Same order on the bottom:

x_{2} - x_{1} = 3 - (- 1) = 4

m = \frac{- 8}{4}

Answer

m = - 2

Why this works

Sketching the two points roughly before computing gives you a free sanity check: if the points visibly fall left to right, the gradient must come out negative.

Writing the equation: start from a point and a gradient

Given a gradient $m$ and any point $(x_{1}, y_{1})$ on the line, the equation is $y - y_{1} = m (x - x_{1})$ . This is the form to reach for by default: it takes exactly the information most questions give you, with no rearranging first.

The familiar $y = m x + c$ is what you get after expanding and tidying. It is a good final form because the intercept is visible, but as a starting point it forces an extra step to find $c$ .

For two points, find the gradient first, then use either point — both give the same line, so pick whichever has friendlier numbers.

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

Point-gradient form: use this first.

y = m x + c

Gradient-intercept form: useful as an answer.

a x + b y + c = 0

The tidy form OCR often asks for, with integer coefficients.

1Point-gradient:

y - (- 1) = 3 (x - 2)

, so

y + 1 = 3 x - 6

2Rearrange to zero:

0 = 3 x - y - 7

Answer

3 x - y - 7 = 0

Tip — When the form $a x + b y + c = 0$ is requested, clear all fractions and give integer coefficients. Leaving a fraction in usually costs the final mark.

Parallel and perpendicular

Parallel lines have equal gradients — same steepness, never meeting. Perpendicular lines have gradients whose product is $- 1$ , so each is the $negative reciprocal$ of the other: flip the fraction and change the sign.

Two special cases sit outside the rule. A horizontal line ( $m = 0$ ) is perpendicular to a vertical line (no gradient), and the product condition cannot express that because one gradient does not exist.

m_{1} = m_{2}

Parallel.

m_{1} m_{2} = - 1

Perpendicular.

m_{2} = - \frac{1}{m _{1}}

The negative reciprocal, rearranged.

1The given gradient is

\frac{2}{3}

2Negative reciprocal: flip to

\frac{3}{2}

, change sign to

- \frac{3}{2}

3Point-gradient:

y - 1 = - \frac{3}{2} (x - 4)

4Expand:

y - 1 = - \frac{3}{2} x + 6

Answer

y = - \frac{3}{2} x + 7

Tip — Do both operations. Turning $\frac{2}{3}$ into $\frac{3}{2}$ without the minus, or into $- \frac{2}{3}$ without flipping, are the two half-done versions of this step.

Midpoint, distance and intersection

The midpoint is the average of the coordinates. The distance is Pythagoras applied to the horizontal and vertical gaps — which is why it is worth leaving in surd form when the question wants exact values.

Two lines intersect where their equations hold simultaneously, so solving them as a pair gives the point. Parallel lines produce a contradiction instead, which is the algebra telling you they never meet.

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

Midpoint: average each coordinate.

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

Distance: Pythagoras on the gaps.

1Horizontal gap:

3 - (- 2) = 5

. Vertical gap:

5 - 1 = 4

d = 5^{2} + 4^{2} = 41

341 is prime, so the surd will not simplify.

Answer

41

Why this works

The distance formula squares both gaps, so their signs are irrelevant — you cannot get a negative length by subtracting in the wrong order. Gradient, by contrast, depends on the order entirely.

Think like an examiner

Read the requested form carefully. " $a x + b y + c = 0$ with integer coefficients" means fractions must be cleared and the answer set equal to zero.
For perpendicular gradients, show the negative reciprocal step explicitly — it is a credited line.
Leave distances as surds when exact answers are asked for; $41$ is the answer, not 6.4.
Straight-line work often appears as an early part of a circle or calculus question, where a slip here costs every later mark.

Common misconceptions

Computing

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

Subtract in the same order top and bottom or the sign inverts.

Taking the perpendicular gradient of

\frac{3}{4}

- \frac{3}{4}

Flip and negate:

- \frac{4}{3}

. Check by multiplying — it must give

- 1

Leaving

y = \frac{2}{5} x + 3

when

a x + b y + c = 0

with integers was requested.

Multiply through by 5 and rearrange:

2 x - 5 y + 15 = 0

Writing a vertical line as

y = k

Vertical lines are

x = k

and have no gradient. Horizontal lines are

y = k

with gradient 0.

Straight-line formulae

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

Gradient from two points.

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

Point-gradient form.

m_{1} m_{2} = - 1

Perpendicular condition.

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

Midpoint.

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

Distance between points.

Stretch yourself

Triangle $A B C$ has vertices $A (1, 2)$ , $B (5, 4)$ and $C (3, 8)$ . Show that the triangle has a right angle, and find the equation of the line through $A$ perpendicular to $B C$ .

Hint — Compute all three gradients and look for a pair whose product is $- 1$ .

Questions students ask

Why prefer

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

over

y = m x + c

Because it takes exactly what questions give you — a point and a gradient — with no intermediate step. Using

y = m x + c

means substituting to find

c

first, which is an extra opportunity to slip.

What is the gradient of a vertical line?

It has none. The change in

x

is zero, so the fraction is undefined. Vertical lines are written

x = k

and sit outside the gradient formulae entirely.

Does it matter which point I use for the equation?

No. Both points lie on the line, so either produces the same equation after simplifying. Choose the one with smaller or non-negative numbers.

How do I know two lines are parallel rather than the same line?

Equal gradients make them parallel; if they also share a point they are identical. Solving them simultaneously distinguishes the cases: parallel gives a contradiction, identical gives an identity.

Key takeaways

Gradient is rise over run, subtracted consistently in both numerator and denominator.
Start from $y - y_{1} = m (x - x_{1})$ ; rearrange to whatever form is requested.
Parallel means equal gradients; perpendicular means the product is $- 1$ .
Midpoint averages coordinates; distance is Pythagoras and often stays a surd.
Vertical lines are $x = k$ and have no gradient.

How this fits the course

Build on

Expanding Brackets

Circles

Leads to

Circles

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Real past-paper questions on Straight Lines, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

Practise this topic

Straight Lines

Gradient: the number that defines the line

Writing the equation: start from a point and a gradient

Parallel and perpendicular

Midpoint, distance and intersection

Practise this topic

Related topics

Straight Lines

Gradient: the number that defines the line

Writing the equation: start from a point and a gradient

Parallel and perpendicular

Midpoint, distance and intersection

Practise this topic

Related topics