Pure 3.1 - Equation of a straight line | Edexcel A-Level Mathematics

A straight-line equation describes every point on a line with just a few constants. You will connect those constants to gradients, intercepts and perpendicular directions, then build and interpret straight-line models.

Gradient, intercept and equation forms

Can the Celsius and Fahrenheit readings ever be the same number? Their conversion graph is a straight line, but its intercept is not zero. We will build that line and find the meeting temperature by the end.

Start with a simpler line: from $(2, 2)$ to $(6, 4)$ , moving $4$ units right raises the height by $2$ . That is a rise of $\frac12$ for every unit right.

You already know

A coordinate pair gives horizontal position first and vertical position second. Substitution checks whether a point lies on a graph: $(2, 2)$ satisfies $y=\frac12x+1$ , because $2=\frac12(2)+1$ .

The same rise-to-run ratio applies between any two distinct points on a non-vertical straight line.

Gradient and intercept

The gradient is the change in $y$ divided by the change in $x$ . The $y$ -intercept is the value of $y$ at $x=0$ . For example, $y=\frac12x+1$ has gradient $\frac12$ and intercept $1$ . A vertical line has no finite gradient.

The picture links the ratio to the graph. A negative gradient descends as you move right; zero gradient gives a horizontal line.

[DIAGRAM: asset_slug: p1_3_1_understand_and_use_the_equation_of_a_straight_line_in_the_x_y_plane__diagram_01; asset_name: Gradient as rise over run; caption: A rise of 2 for a run of 4 gives gradient one half; description: Coordinate graph of y equals one half x plus 1 with points (2, 2) and (6, 4), a rise-run triangle and y-intercept (0, 1)]

Gradient and equation forms

For distinct points with $x_2\ne x_1$ ,

m=\frac{y_2-y_1}{x_2-x_1},\qquad y=mx+c

In the second form, $c$ is the $y$ -intercept. The general form is $ax+by+c=0$ , where $a$ and $b$ are not both zero. Its constant $c$ is generally not the intercept. Not in the formula booklet: learn it.

To read a finite gradient from general form, make $y$ the subject first.

Worked example 1: reading an expanded equation

Question. Find the gradient and $y$ -intercept of $4x-2y+6=0$ .

Plan. Rearrange into $y=mx+c$ , then read the two constants.

\begin{aligned} -2y&=-4x-6 &&\text{move the other terms}\\ y&=2x+3 &&\text{divide every term by }-2 \end{aligned}

Why this step? The coefficient of $x$ is the gradient only when the coefficient of $y$ is $1$ .

Answer. Gradient $2$ , $y$ -intercept $3$ .

Check. At $x=0$ , the original equation gives $-2y+6=0$ , so $y=3$ .

Change only the sign of the $y$ term. This reverses both signs after rearrangement, so watch the whole equation.

Worked example 2: a changed sign

Question. Find the gradient and $y$ -intercept of $4x+2y+6=0$ .

Plan. Isolate $y$ before interpreting the coefficients.

\begin{aligned} 2y&=-4x-6 &&\text{subtract }4x+6\\ y&=-2x-3 &&\text{divide by }2 \end{aligned}

Why this step? Reading $4$ straight from the original equation would ignore the rearrangement.

Answer. Gradient $-2$ , $y$ -intercept $-3$ .

Check. The points $(0, -3)$ and $(1, -5)$ both satisfy the original equation and give gradient $-2$ .

The exceptional lines still fit general form: $x=4$ is vertical and $y=-1$ is horizontal. Only the second has a finite gradient.

Watch out

Wrong: “ $3x+2y-8=0$ has intercept $-8$ .” Right: $y=-\frac32x+4$ , so its intercept is $4$ . The constant in general form is not automatically the intercept.

Try the same reading method with one changed constant.

Finding a line through points

Two distinct points fix exactly one straight line. For a non-vertical line, find its gradient and attach it to one of the points.

Take $A(2, -1)$ and $B(8, 3)$ . Their horizontal separation is $6$ , so the gradient calculation is possible. Any other point $(x, y)$ on their line must share that same gradient from $A$ .

Point-gradient form

A non-vertical line through $(x_1, y_1)$ with gradient $m$ has equation

y-y_1=m(x-x_1)

It follows by multiplying $\frac{y-y_1}{x-x_1}=m$ by $x-x_1$ ; the resulting equation also includes the known point. Not in the formula booklet: learn it.

Either known point works. Choose the one whose signs make the substitution easiest.

Worked example 3: two points to general form

Question. Find the line through $A(2, -1)$ and $B(8, 3)$ in the form $ax+by+c=0$ .

Plan. Calculate the gradient, use $A$ , then clear the denominator.

\begin{aligned} m&=\frac{3-(-1)}{8-2}=\frac23 &&\text{rise divided by run}\\ y+1&=\frac23(x-2) &&\text{use }A\\ 3y+3&=2x-4 &&\text{multiply by }3 \end{aligned}

Why this step? Point-gradient form uses both the direction and the position; a gradient alone describes many parallel lines.

Answer. $2x-3y-7=0$ .

Check. Substitution gives $4+3-7=0$ at $A$ and $16-9-7=0$ at $B$ .

Change only the height of $B$ , from $3$ to $-5$ . The line now falls, but the method stays put.

Worked example 4: a negative gradient

Question. Find the line through $A(2, -1)$ and $B(8, -5)$ in general form.

Plan. Keep the subtraction order consistent, then use $A$ .

\begin{aligned} m&=\frac{-5-(-1)}{8-2}=-\frac23 &&\text{same point order}\\ y+1&=-\frac23(x-2) &&\text{point-gradient form}\\ 3y+3&=-2x+4 &&\text{clear the fraction} \end{aligned}

Why this step? Both coordinate differences must run from the same first point to the same second point.

Answer. $2x+3y-1=0$ .

Check. At $A$ , $4-3-1=0$ ; at $B$ , $16-15-1=0$ .

If the two $x$ -coordinates agree, stop before dividing: points such as $(2, -1)$ and $(2, 5)$ give the vertical line $x=2$ . Two identical points do not determine a unique line.

Watch out

Wrong: $m=\frac{-5+1}{2-8}=\frac23$ , mixing point orders. Right: $m=\frac{-5+1}{8-2}=-\frac23$ . A quick sketch would also expose the wrong sign.

Now change the second point's height again and keep the requested final form.

Parallel and perpendicular lines

Sliding a line without rotating it keeps its gradient. Turning its direction through a right angle swaps the roles of rise and run and reverses one sign.

For instance, direction steps $(4, 3)$ and $(3, -4)$ give gradients $\frac34$ and $-\frac43$ . Their product is $-1$ .

[DIAGRAM: asset_slug: p1_3_1_understand_and_use_the_equation_of_a_straight_line_in_the_x_y_plane__diagram_02; asset_name: Parallel and perpendicular directions; caption: Parallel lines share a gradient while perpendicular finite gradients multiply to minus one; description: Two panels show parallel lines of gradient three quarters and perpendicular lines of gradients three quarters and minus four thirds]

Parallel and perpendicular gradients

For distinct non-vertical parallel lines, $m'=m$ . For perpendicular lines with finite gradients,

mm'=-1,\qquad m'=-\frac1m\quad(m\ne0)

A horizontal line is perpendicular to a vertical line; handle that case directly. Not in the formula booklet: learn it.

First identify the given gradient, then decide which relationship the question needs. Equal gradients can also describe the very same line, so check the point if distinct parallel lines are required.

Worked example 5: perpendicular to a given line

Question. Find the line perpendicular to $4x+3y-6=0$ through $(6, -2)$ .

Plan. Read the original gradient, take its negative reciprocal and use the new point.

\begin{aligned} y&=-\frac43x+2 &&\text{isolate }y\\ m'&=\frac34 &&\text{negative reciprocal}\\ y+2&=\frac34(x-6) &&\text{use the required point} \end{aligned}

Why this step? The perpendicular line needs the reciprocal gradient, not the original direction.

Answer. $3x-4y-26=0$ .

Check. $(-\frac43)(\frac34)=-1$ , and $18+8-26=0$ at the point.

Change only the relationship from perpendicular to parallel. The same original line and point now lead to a different answer.

Worked example 6: parallel instead

Question. Find the line parallel to $4x+3y-6=0$ through $(6, -2)$ .

Plan. Keep the gradient and use point-gradient form.

\begin{aligned} m'&=-\frac43 &&\text{parallel direction}\\ y+2&=-\frac43(x-6) &&\text{use }(6,-2)\\ 3y+6&=-4x+24 &&\text{clear the denominator} \end{aligned}

Why this step? Changing “perpendicular” to “parallel” changes the direction rule, not the line-through-a-point method.

Answer. $4x+3y-18=0$ .

Check. Both lines have gradient $-\frac43$ ; the new equation gives $24-6-18=0$ .

For a vertical starting line such as $x=4$ , a parallel line through $(6, -2)$ is $x=6$ , and a perpendicular one is $y=-2$ .

Watch out

Wrong: a perpendicular to gradient $-\frac43$ has gradient $\frac43$ . Right: it has gradient $\frac34$ . Change the sign and invert the fraction.

Change the point and retain the parallel requirement.

Straight-line models

A sensor that gives $1.2\,\mathrm V$ at $10^\circ\mathrm C$ and $2.0\,\mathrm V$ at $30^\circ\mathrm C$ rises by $0.8\,\mathrm V$ over $20^\circ\mathrm C$ . Assuming a straight response between these readings turns that rate into the gradient of a model.

Straight-line model

A straight-line model assumes one quantity changes at a constant rate with another: $y=mx+c$ . The gradient is that rate and the intercept is the predicted value at $x=0$ . Two measurements fit a line exactly but do not prove the actual relationship is linear.

A conversion scale and a measured sensor response can share the same algebra while having different reliability. A calibration range belongs with its model.

[DIAGRAM: asset_slug: p1_3_1_understand_and_use_the_equation_of_a_straight_line_in_the_x_y_plane__diagram_03; asset_name: Straight-line models and units; caption: The gradient has units of output per unit input; description: Celsius-Fahrenheit conversion through (0, 32) and (100, 212), beside a distance-time model s equals 8t plus 2 with intercept 2 metres and gradient 8 metres per second]

Worked example 7: fitting a sensor model

Question. Model the sensor readings above as $V=aT+b$ . Interpret $a$ .

Plan. Divide the output change by the temperature change, then use one reading for $b$ .

\begin{aligned} a&=\frac{2.0-1.2}{30-10}=0.04 &&\text{volts per degree Celsius}\\ 1.2&=0.04(10)+b &&\text{use the first reading}\\ b&=0.8 &&\text{subtract }0.4 \end{aligned}

Why this step? The intercept must be found from a point; it is not the first measured voltage unless its temperature is zero.

Answer. $V=0.04T+0.8$ . The voltage rises by $0.04\,\mathrm V$ per $1^\circ\mathrm C$ , under the model.

Check. At $T=30$ , $V=1.2+0.8=2.0$ . The readings support use between $10^\circ\mathrm C$ and $30^\circ\mathrm C$ .

Change only the second voltage, to $2.4\,\mathrm V$ . The intercept must be recalculated as well as the gradient.

Worked example 8: a changed calibration reading

Question. Fit $V=aT+b$ through $(10, 1.2)$ and $(30, 2.4)$ , and predict $V$ at $20^\circ\mathrm C$ .

Plan. Fit both constants, then interpolate inside the measured interval.

\begin{aligned} a&=\frac{2.4-1.2}{30-10}=0.06 &&\text{change divided by change}\\ b&=1.2-0.06(10)=0.6 &&\text{use the first reading}\\ V(20)&=0.06(20)+0.6=1.8 &&\text{substitute the temperature} \end{aligned}

Why this step? A new rate through a fixed measured point usually changes the intercept too.

Answer. $V=0.06T+0.6$ , predicting $1.8\,\mathrm V$ .

Check. $20$ is halfway between $10$ and $30$ ; $1.8$ is halfway between $1.2$ and $2.4$ .

The earlier distance model $s=8t+2$ , with $s$ in metres and $t$ in seconds, gives constant speed $8\,\mathrm{m\,s^{-1}}$ and initial distance $2\,\mathrm m$ . In contrast, readings $(3, 17)$ and $(9, 53)$ give $s=6t-1$ : use it over the observed interval, because its negative prediction at $t=0$ cannot be a distance.

Watch out

Wrong: “Two correct predictions prove the sensor is linear at every temperature.” Right: the line fits the readings; extending beyond the measured range requires additional evidence. This extension is extrapolation.

Change only the second calibration voltage again; the prediction is still an interpolation.

Choosing the method

The information in a problem tells you where to begin. Two points determine a direction; a perpendicular condition changes that direction; a supplied point locates the required line.

For $A(0, 4)$ , $B(6, 1)$ , the line $AB$ has gradient $-\frac12$ . A perpendicular through $B$ therefore has gradient $2$ , giving $y-1=2(x-6)$ , or $y=2x-11$ . The point belongs to the new line; the direction comes from the old one.

Given information	First mathematical decision
A point and a gradient	Use point-gradient form
Two distinct points	Find the gradient, or recognise a vertical line
A parallel or perpendicular condition	Obtain the given gradient before choosing the new one
A conversion model and equal readings	Set the two measured numbers equal and solve

The opening puzzle uses the last row. A linear conversion through $(C, F)=(0, 32)$ and $(100, 212)$ has gradient $\frac{212-32}{100}=\frac95$ , so

F=\frac95C+32

At equal readings set $F=C$ :

C=\frac95C+32\quad\Rightarrow\quad-\frac45C=32\quad\Rightarrow\quad C=-40

The surprising fact is that $-40^\circ\mathrm C$ and $-40^\circ\mathrm F$ are the same temperature. The algebra verifies it: $\frac95(-40)+32=-40$ . Unlike a sensor extrapolation, this follows from the conversion rule itself.

Key idea

Choose a starting equation from the information you actually have, then check that the result satisfies every condition in the question.

A useful check can catch a wrong choice before the algebra becomes long. Decide which line below has the first error.

Quick Recap

Read gradient and intercept after rearranging to $y=mx+c$ .
Use two points for the gradient, then one point in $y-y_1=m(x-x_1)$ .
Parallel gradients agree; perpendicular finite gradients multiply to $-1$ , with horizontal/vertical cases handled directly.
Interpret a model's rate, intercept and valid range in context.

The Celsius and Fahrenheit readings agree at $-40$ , found by intersecting the conversion line with $F=C$ .

Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 3.1 - Equation of a straight line

Gradient, intercept and equation forms

Finding a line through points

Parallel and perpendicular lines

Straight-line models

Choosing the method

Quick Recap

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