C1 - Straight lines and linear models | AQA A-Level Mathematics

Find line equations, use coordinates to justify lengths and angles, and build straight-line models with meaningful rates and valid input ranges. You will combine these ideas to prove the shape of a quadrilateral without trusting the appearance of a sketch.

Gradient and line equations

How can you prove that a tilted quadrilateral is a square when none of its sides is horizontal? Start with a small piece of the puzzle: moving from $(0,0)$ to $(2,1)$ means moving $2$ across and $1$ up, so the side's gradient is $1/2$ . Coordinates give exact evidence about direction, even when a drawing is distorted.

You already know

A point $(x,y)$ has horizontal coordinate $x$ and vertical coordinate $y$ . Substituting both coordinates into an equation tests whether the point lies on its graph. Solving two line equations simultaneously finds their intersection.

The gradient measures change in the vertical coordinate per unit change in the horizontal coordinate. Subtract coordinates in the same order; reversing both differences leaves their ratio unchanged.

Gradient and line equations

For distinct points $(x_1,y_1)$ and $(x_2,y_2)$ , with $x_1\ne x_2$ ,

m=\frac{y_2-y_1}{x_2-x_1}

The line through $(x_1,y_1)$ with finite gradient $m$ has equation

y-y_1=m(x-x_1)

A general straight-line equation is $ax+by+c=0$ , where $a,b$ are not both zero. If $b\ne0$ , its gradient is $-a/b$ .

Not in the formula booklet: learn it.

The point-gradient form comes from the constant ratio $(y-y_1)/(x-x_1)=m$ . Multiplying by $x-x_1$ gives an equation that also works at the known point. Rearranging to $y=mx+c$ displays the gradient and vertical intercept. Multiplying a whole line equation by any nonzero constant leaves the line unchanged.

Worked example 1: a line through two points

Question. Find the line through $A(-2,5)$ and $B(4,-7)$ , in the form $ax+by+c=0$ .

Plan. Calculate the gradient, use one known point, then rearrange into the requested form.

\begin{aligned} m&=\frac{-7-5}{4-(-2)}=-2 && \text{same subtraction order} \\ y-5&=-2(x+2) && \text{use } A \\ y&=-2x+1 && \text{expand and simplify} \\ 2x+y-1&=0 && \text{requested form} \end{aligned}

Why this step? A gradient fixes direction but allows many parallel lines; using $A$ fixes the position.

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

Check. At $B$ , $2(4)-7-1=0$ ; at $A$ , $-4+5-1=0$ .

Now change just one feature: the two points have the same horizontal coordinate. The general line form still works, but the gradient calculation does not.

Worked example 2: the vertical boundary case

Question. Find the line through $(3,-2)$ and $(3,5)$ , in the form $ax+by+c=0$ . State its gradient.

Plan. Identify the coordinate that is constant instead of dividing by zero.

\begin{aligned} \Delta x&=3-3=0 && \text{no horizontal change} \\ x&=3 && \text{both points have this coordinate} \\ x-3&=0 && \text{general line form} \end{aligned}

Why this step? A vertical line cannot be written as $y=mx+c$ with a finite $m$ ; its equation fixes $x$ , not $y$ .

Answer. $x-3=0$ . Its gradient is undefined.

Check. Both supplied points have $x=3$ , and every point with $x=3$ is on the same vertical line.

A horizontal line instead has equation $y=k$ and gradient zero. These horizontal and vertical forms are also not in the formula booklet: learn them. In AQA questions, any correct simplified form is acceptable unless a form is specified.

Watch out

Wrong: the gradient of $4x+3y-6=0$ is $4/3$ . Right: $y=-\frac43x+2$ , so it is $-4/3$ . Rearranging makes the sign visible.

Change the boundary case once more: this time the vertical coordinate is constant.

Midpoints and distances

Coordinates also measure position and length: the midpoint is halfway in each coordinate, while the segment's length comes from a right triangle.

For $(-1,2)$ and $(5,10)$ , the horizontal change is $6$ and the vertical change is $8$ . The midpoint lies $3$ across and $4$ up from the first point; Pythagoras gives length $10$ .

Midpoint and distance

For $A(x_1,y_1)$ and $B(x_2,y_2)$ ,

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

AB^2=(x_2-x_1)^2+(y_2-y_1)^2 \qquad AB=\sqrt{AB^2}

A length uses the nonnegative square root. These are coordinate forms of averaging and Pythagoras.

Not in the formula booklet: learn it.

Compare squared lengths when you only need to decide whether lengths are equal. This avoids unnecessary surds without losing information, because lengths are nonnegative.

Worked example 3: an exact segment length

Question. Find the midpoint and length of the segment joining $A(-1,2)$ to $B(5,10)$ .

Plan. Average each coordinate and use Pythagoras on the coordinate changes.

\begin{aligned} M&=\left(\frac{-1+5}{2},\frac{2+10}{2}\right)=(2,6) && \text{coordinate averages} \\ AB^2&=(5+1)^2+(10-2)^2=100 && \text{Pythagoras} \\ AB&=10 && \text{positive length} \end{aligned}

Why this step? Horizontal and vertical changes are perpendicular, so they are the two shorter sides of a right triangle.

Answer. Midpoint $(2,6)$ , length $10$ .

Check. From $M$ to either endpoint, the squared distance is $3^2+4^2=25$ , one quarter of $AB^2$ .

The next example changes the unknown: one endpoint is missing, while the midpoint is known. Doubling the midpoint reverses the averaging operation.

Worked example 4: recovering an endpoint

Question. The midpoint of $AB$ is $M(2,6)$ and $A(-1,2)$ . Find $B$ and the length $AB$ .

Plan. Solve the two midpoint equations, then calculate the distance.

\begin{aligned} \frac{-1+x_B}{2}&=2 && \text{horizontal midpoint} \\ x_B&=5 && \text{double, then add } 1 \\ \frac{2+y_B}{2}&=6 && \text{vertical midpoint} \\ y_B&=10 && \text{double, then subtract } 2 \\ AB&=\sqrt{6^2+8^2}=10 && \text{distance formula} \end{aligned}

Why this step? The midpoint is the average of the endpoints, so it determines the missing endpoint uniquely once the other is known.

Answer. $B(5,10)$ , and $AB=10$ .

Check. $A$ to $M$ and $M$ to $B$ both have change $(3,4)$ .

You can use this reversal later to find missing vertices and circle diameters.

Watch out

Wrong: average an $x$ coordinate with a $y$ coordinate to find a midpoint. Right: average the two horizontal coordinates together and the two vertical coordinates together. For a missing endpoint, reverse each average separately.

The next question varies the coordinates while keeping the same midpoint reversal.

Parallel and perpendicular lines

A gradient tells you a line's direction; comparing gradients tells you whether two directions agree or meet at a right angle.

Two distinct nonvertical lines with gradient $2$ are parallel. With gradients $2$ and $-1/2$ , they are perpendicular. Equal gradients can also describe the same line, so check the intercepts when you need distinct parallel lines. Distinct vertical lines are parallel too; their gradients are undefined, so compare their equations $x=h$ instead.

Direction tests

Distinct nonvertical lines are parallel exactly when $m_1=m_2$ .

For two lines with finite gradients, perpendicularity is equivalent to

m_1m_2=-1

Thus the perpendicular gradient to a nonzero gradient $m$ is $-1/m$ . A horizontal line and a vertical line are perpendicular; the finite-gradient product test does not apply to that pair.

Not in the formula booklet: learn it.

The negative reciprocal has a geometric reason. A direction with horizontal and vertical changes $(u,v)$ turns through a right angle to $(-v,u)$ . When $u,v\ne0$ , their gradients are $v/u$ and $-u/v$ , whose product is $-1$ .

Worked example 5: a parallel line

Question. Find the line through $(3,-1)$ parallel to $4x+3y-6=0$ .

Plan. Read the given gradient, retain it, and use the supplied point.

\begin{aligned} y&=-\frac43x+2 && \text{rearrange the given line} \\ y+1&=-\frac43(x-3) && \text{same gradient through the point} \\ 4x+3y-9&=0 && \text{simplify} \end{aligned}

Why this step? Parallel lines have the same direction; changing the intercept moves the line without turning it.

Answer. $4x+3y-9=0$ .

Check. The gradients agree, the intercepts $2$ and $3$ differ, and $4(3)+3(-1)-9=0$ .

Change only the required relationship from parallel to perpendicular. Now keep the point but change the gradient to its negative reciprocal.

Worked example 6: a perpendicular line

Question. Find the line through $(3,-1)$ perpendicular to $4x+3y-6=0$ .

Plan. Replace the given gradient by its negative reciprocal, then use point-gradient form.

\begin{aligned} m&=-\frac43 && \text{given direction} \\ m_{\perp}&=\frac34 && \text{negative reciprocal} \\ y+1&=\frac34(x-3) && \text{through } (3,-1) \\ 3x-4y-13&=0 && \text{simplify} \end{aligned}

Why this step? Merely changing the sign would give $4/3$ , whose product with $-4/3$ is not $-1$ .

Answer. $3x-4y-13=0$ .

Check. $(-4/3)(3/4)=-1$ , and $9+4-13=0$ at the required point.

If the given line is horizontal, use a vertical answer directly. No fraction involving an undefined gradient is needed.

Watch out

Wrong: a line perpendicular to $y=7$ has gradient $-1/0$ . Right: $y=7$ is horizontal, so its perpendicular through $(-4,9)$ is the vertical line $x=-4$ .

The next question changes the given line's gradient while retaining the perpendicular construction.

Coordinate quadrilateral reasoning

To identify a quadrilateral, calculate enough properties to force the conclusion. A shape looking square, or having just one familiar property, is insufficient.

For a simple, nondegenerate quadrilateral $ABCD$ , with vertices given in order, these sufficient conditions are useful. They are not supplied in the formula booklet: learn the properties and justify their use.

Shape	Sufficient coordinate evidence
Parallelogram	Both opposite side pairs parallel, or diagonals with the same midpoint
Rectangle	A parallelogram with one right angle
Rhombus	A parallelogram with two adjacent sides equal
Square	A rectangle with two adjacent sides equal
Kite	Two equal adjacent side pairs, such as $AB=AD$ and $BC=CD$
Trapezium	A pair of opposite sides parallel; show the other pair is not parallel if exactly one pair is required

These are necessary as well as sufficient under the inclusive definitions used here: for example, every rectangle is a parallelogram with a right angle, and that combination forces a rectangle. Squares also satisfy the rectangle and rhombus definitions. To distinguish a kite from a rhombus, show the two pair-lengths differ. Equal diagonals alone do not prove a rectangle; perpendicular diagonals alone do not prove a square.

Worked example 7: a tilted square

Question. Prove that consecutive vertices $A(0,0),B(2,1),C(1,3),D(-1,2)$ form a square.

Plan. Establish a parallelogram, a right angle and equal adjacent lengths.

\begin{aligned} m_{AB}=m_{CD}&=\frac12 && \text{opposite sides agree} \\ m_{BC}=m_{DA}&=-2 && \text{other opposite sides agree} \\ \frac12(-2)&=-1 && \text{adjacent sides perpendicular} \\ AB^2&=2^2+1^2=5 && \text{first squared length} \\ BC^2&=(-1)^2+2^2=5 && \text{adjacent squared length} \end{aligned}

Why this step? Parallel opposite sides give a parallelogram; the right angle makes it a rectangle; equal adjacent lengths then make it a square.

Answer. $ABCD$ is a square.

Check. The diagonal midpoints are both $(1/2,3/2)$ , independently confirming the parallelogram structure.

Here is a pleasing surprise: all four vertices have integer coordinates, yet every side has length $\sqrt5$ . A square need not line up with the grid to be exact. The squared-length calculation above proves it.

The next example changes only one decisive feature: the adjacent lengths are unequal. The direction tests still prove a rectangle.

Worked example 8: a rectangle that is not a square

Question. Prove that $A(1,1),B(5,3),C(4,5),D(0,3)$ , in that order, form a rectangle but not a square.

Plan. Repeat the direction tests, then compare squared adjacent lengths.

\begin{aligned} m_{AB}=m_{CD}&=\frac12 && \text{first opposite pair} \\ m_{BC}=m_{DA}&=-2 && \text{second opposite pair} \\ m_{AB}m_{BC}&=-1 && \text{one right angle} \\ AB^2&=4^2+2^2=20 && \text{longer side} \\ BC^2&=(-1)^2+2^2=5 && \text{shorter side} \end{aligned}

Why this step? The unequal lengths rule out a square only after the direction evidence has established the rectangle.

Answer. It is a rectangle, and it is not a square because $AB\ne BC$ .

Check. Both diagonals have midpoint $(5/2,3)$ , and both have squared length $25$ , consistent with a rectangle.

The equal axis scales show the geometry faithfully, but the proof comes from the calculations.

The rectangle with vertices A (1, 1), B (5, 3), C (4, 5) and D (0, 3), plotted on equal scales

One suggestive property rarely settles the classification. Length evidence and direction evidence answer different questions.

Watch out

Wrong: four equal sides prove a square. Right: they prove a rhombus, which may have no right angle. Establish perpendicular adjacent sides as well. Equal diagonals alone are also insufficient to prove a rectangle.

For the next rectangle, change the coordinates again. You still need both a direction argument and a length comparison.

Constructing linear models

A straight-line model assumes a constant change in one quantity for each unit increase in another, over a stated range.

A hire charge rising from £19 for two hours to £39 for seven hours increases by £20 over five hours. A constant rate therefore means £4 per hour. The intercept accounts for the rest of the charge.

Linear model

A model $y=mx+c$ has rate $m$ , the output change per input unit, and intercept $c$ , the predicted output at $x=0$ . For example, a fixed charge plus a constant hourly rate gives a linear cost model. A price that changes its hourly rate after a threshold needs a different rule or separate pieces.

Define variables and units before fitting a line. Time may vary continuously, while a number of parcels is restricted to whole numbers; state whichever domain the context permits. A model can fit two observations exactly without being valid for every input. Interpolation uses inputs between observations; extrapolation goes beyond them and needs further justification. The linear model form and the coefficient calculations below are not in the formula booklet: learn the method.

Worked example 9: a hire charge

Question. A company charges a fixed fee plus a constant hourly fee for hires from one to eight hours. Two hours cost £19 and seven hours cost £39. Find a cost model, interpret its coefficients and predict the cost for five hours.

Plan. Use the change in cost to find the hourly rate, then substitute one observation to find the fixed fee.

\begin{aligned} m&=\frac{39-19}{7-2}=4 && \text{pounds per hour} \\ 19&=4(2)+c && \text{use the first quote} \\ c&=11 && \text{fixed fee} \\ C(5)&=4(5)+11=31 && \text{use the fitted rule} \end{aligned}

Why this step? Taking differences removes the fixed fee, leaving just the cost of the extra hours.

Answer. $C=4h+11$ , for $1\le h\le8$ , with $h$ in hours and $C$ in pounds. The rate is £4 per hour, the fixed fee is £11, and five hours cost £31.

Check. $C(7)=28+11=39$ , matching the second quote.

The next example changes the sign of the rate: the quantity decreases at a constant rate. A negative gradient is meaningful, provided you interpret the variables correctly.

Worked example 10: a draining tank

Question. A tank contains $230\,\mathrm L$ after two minutes of draining and $140\,\mathrm L$ after eight minutes. Assume a constant draining rate until it empties. Find $V$ in terms of elapsed time $t$ , and the time at which the model predicts an empty tank.

Plan. Fit the line from the observations, then solve $V=0$ to locate the model's end.

\begin{aligned} m&=\frac{140-230}{8-2}=-15 && \text{litres per minute} \\ 230&=-15(2)+c && \text{use one observation} \\ V&=260-15t && \text{volume model} \\ 0&=260-15t && \text{empty tank} \\ t&=\frac{52}{3} && \text{solve for time} \end{aligned}

Why this step? Negative volume is impossible, so the algebraic line must stop at the emptying event.

Answer. $V=260-15t$ , for $0\le t\le52/3$ , with volume in litres and time in minutes. The predicted emptying time is $17\frac13$ minutes. The intercept is the model's initial $260\,\mathrm L$ .

Check. At eight minutes, $260-120=140$ . At the endpoint, $260-15(52/3)=0$ .

The emptying time is extrapolated beyond the measured times: it is a conditional prediction of constant flow. Real drainage may slow as the level falls, so another observation would test that assumption. A temperature-conversion line gives another interpretation: its gradient changes the scale, while its intercept shifts the zero point.

Watch out

Wrong: two matching readings prove the rate stays constant. Right: the readings determine the fitted line; the constant-rate assumption and the valid range still need contextual support.

The next tank has a different constant decrease. Keep the event that ends the model in your answer.

Quick Recap

A finite gradient and a point give $y-y_1=m(x-x_1)$ ; the general form also includes vertical lines.
Use coordinate averages for midpoints and Pythagoras for lengths.
Equal gradients establish parallel directions; finite perpendicular gradients multiply to $-1$ .
Prove quadrilaterals using sufficient direction and length conditions, with vertices in order.
A linear model needs defined variables, a rate, an intercept and a justified input range.

The tilted quadrilateral from the opening question is proved square by parallel opposite sides, one right angle and equal adjacent lengths; no judgement by eye is needed.

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C1 - Straight lines and linear models

Gradient and line equations

Midpoints and distances

Parallel and perpendicular lines

Coordinate quadrilateral reasoning

Constructing linear models

Quick Recap

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