2.1 Equations of straight lines

A straight line is the simplest graph of a relationship between two variables: one that changes at a constant rate. Lines model situations such as a taxi fare that rises by a fixed amount per kilometre or the conversion between temperature scales, and they reappear throughout calculus as tangents and normals to curves. Almost every topic in the course uses the equation of a line somewhere, so fluency with it saves time everywhere else.

Every non-vertical line is completely described by two numbers: how steep it is and where it crosses the $y$-axis. Working with lines mostly means finding those two numbers from different kinds of information and then using them.

Gradient

The steepness of a line is measured by its gradient (also called the slope). Moving between any two points on the line, the vertical change (the rise) divided by the horizontal change (the run) always gives the same number.

The value does not depend on which two points are chosen. Any two rise–run triangles drawn on the same line have the same angles, so they are similar, and the ratio of their vertical side to their horizontal side is the same. This is exactly what makes a line straight: its gradient is constant everywhere along it.

The sign and size of the gradient

The sign of the gradient shows the direction of the line, read from left to right. Its size shows how steep the line is: a gradient of $5$ is steeper than a gradient of $\frac{1}{2}$, and a gradient of $-5$ is steeper than a gradient of $-\frac{1}{2}$.

A vertical line has no defined gradient because every pair of points on it has $x_2 - x_1 = 0$, and division by zero is undefined. It is wrong to say its gradient is "infinity" or zero.

Reading a gradient from a graph

To find the gradient from a graph, choose two points where the line passes exactly through grid intersections, so that their coordinates can be read without estimating. Then count (or subtract) the rise and the run, taking care with the scales on the two axes, which are often different.

Gradient as a rate of change

When the axes stand for real quantities, the gradient has units: units of $y$ per unit of $x$. If $y$ is a distance in kilometres and $x$ a time in hours, the gradient is a speed in kilometres per hour. A constant gradient means a constant rate of change, which is the defining feature of a linear relationship. The idea is extended to curves in calculus (5.1), where the gradient changes from point to point.

Forms of the equation of a line

A point $(x, y)$ lies on a line exactly when its coordinates satisfy the line's equation. Three forms of the equation appear regularly. Each is convenient for a different purpose, and any one can be rearranged into the others.

Gradient–intercept form

In $y = mx + c$, the coefficient $m$ is the gradient and the constant $c$ is the y-intercept, the value of $y$ where the line crosses the $y$-axis.

The form follows directly from the gradient formula. A line with gradient $m$ passes through $(0, c)$. For any other point $(x, y)$ on it,

Point–gradient form

Often the $y$-intercept is not known, but the gradient $m$ and one point $(x_1, y_1)$ on the line are. If $(x, y)$ is any other point on the line, the gradient between it and $(x_1, y_1)$ must equal $m$:

This is the point–gradient form. It is the quickest route from "a point and a gradient" to an equation, and it is the form used for tangents and normals in calculus (5.4).

When two points are given instead of a point and a gradient, find the gradient first and then use either point in the point–gradient form. Both points give the same line.

General form

The general form $ax + by + d = 0$ puts every term on one side. It is the standard way to present a final answer because the coefficients can always be made integers, and it is the only one of the three forms that can describe every line, vertical lines included.

To find the gradient and $y$-intercept from the general form (with $b \neq 0$), make $y$ the subject:

so the gradient is $-\frac{a}{b}$ and the $y$-intercept is $-\frac{d}{b}$. Rearranging each time is safer than memorising these results.

The same line has many general-form equations: $3x + 2y - 5 = 0$ and $6x + 4y - 10 = 0$ describe the same points. It is usual to divide out any common factor and to make $a$ positive.

Intercepts with the axes

The x-intercept is where the line crosses the $x$-axis. Every point on the $x$-axis has $y = 0$, and every point on the $y$-axis has $x = 0$, which gives a simple method for both intercepts.

Plotting the two intercepts is also the fastest way to draw a line given in general form: find where it meets each axis and join the two points.

Horizontal and vertical lines

A horizontal line has gradient $0$, so $y = mx + c$ becomes $y = c$: every point on it has the same $y$-coordinate. A vertical line has an undefined gradient and cannot be written as $y = mx + c$ at all. Every point on it has the same $x$-coordinate, so its equation is $x = k$ for a constant $k$.

In general form, a horizontal line has $a = 0$ (for example $y - 2 = 0$) and a vertical line has $b = 0$ (for example $x + 3 = 0$). A horizontal line and a vertical line are always perpendicular to each other.

Parallel and perpendicular lines

Parallel lines

Parallel lines run in the same direction and never meet. Two lines run in the same direction exactly when they have the same rise for the same run, which means the same gradient.

For lines in general form, parallel lines have the same ratio $a : b$. So every line parallel to $2x + 5y - 7 = 0$ can be written $2x + 5y + d = 0$ for some constant $d$, which gives a shortcut.

Perpendicular lines

Two lines are perpendicular when they meet at a right angle. Their gradients are linked by a simple rule.

The gradient $-\frac{1}{m_1}$ is called the negative reciprocal of $m_1$: turn the fraction upside down and change its sign. The rule comes from turning a rise–run triangle through a right angle.

Suppose the first line rises $b$ units for a run of $a$ units to the right, so $m_1 = \frac{b}{a}$. Rotating the whole picture through $90°$ anticlockwise turns the run of $a$ to the right into a rise of $a$, and the rise of $b$ into a run of $b$ to the left. The rotated line is perpendicular to the first, and its gradient is

For a "show that" question, every step must be visible and the final sentence must state the conclusion; stopping at $-1$ without saying what it proves leaves the argument unfinished.

Intersection of two lines

The point where two lines cross lies on both of them, so its coordinates satisfy both equations at once. Finding it means solving the two equations simultaneously, by substitution or by elimination. With a graphic display calculator (GDC), the intersection can be found from the graphs or with the simultaneous-equation solver.

Two lines in the plane meet in exactly one point, in no point or in infinitely many points.

When both equations are in general form, elimination is usually quicker. For $2x + 3y = 7$ and $5x - 3y = 14$, adding the equations removes $y$ and gives $7x = 21$, so $x = 3$ and then $3y = 7 - 6$, giving $y = \frac{1}{3}$. The intersection is $\left(3, \frac{1}{3}\right)$.

The perpendicular bisector of a line segment

The perpendicular bisector of a line segment $AB$ cuts it into two equal halves and meets it at right angles. Its key property is that every point on it is equidistant from $A$ and $B$, which is why it is used to find the centre of a circle through given points.

The method uses the midpoint formula (covered in 3.1), $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$, together with the perpendicular gradient rule.

1. Find the midpoint $M$ of $AB$.

2. Find the gradient of $AB$, then the negative reciprocal.

3. Use the point–gradient form with $M$ and the perpendicular gradient.

The perpendicular bisectors of the sides of a triangle all meet at one point, which is equidistant from all three vertices. That point is the centre of the circle through the three vertices.

Lines as linear models

Many real relationships change at a constant rate, at least approximately and over a limited range. A linear model describes such a relationship with a straight line.

Interpreting a model means stating what $m$ and $c$ mean in the context, with units. In a taxi fare $C = 1.6d + 4$, in euros for a journey of $d$ kilometres, the gradient $1.6$ is the cost per kilometre (€1.60 per km) and the intercept $4$ is the fixed charge (€4) paid before the taxi moves. A model is only valid for sensible inputs: here $d \ge 0$, and the line is drawn only for those values.

When a context gives two pairs of values instead of a rate and a starting value, treat them as two points on the line. The temperature scales are a standard example: water freezes at $0\,°\text{C}$, which is $32\,°\text{F}$, and boils (at sea level) at $100\,°\text{C}$, which is $212\,°\text{F}$.

The gradient and the intercept carry the meaning of a linear model, which is a clear case of the course's key concept of modelling: two numbers summarise a whole relationship. The same idea returns in statistics (4.4), where a regression line is a linear model fitted to data that do not lie exactly on a line.

Comparing two linear models

Two linear models can be compared by finding where their lines intersect: on one side of the intersection one model gives the larger value, and on the other side the other does.

Straight lines in calculus

The equation of a line returns in calculus. The tangent to a curve at a point is the line through that point whose gradient equals the gradient of the curve there, and the normal is the line through the same point perpendicular to the tangent. Both are found with the point–gradient form and the rule $m_1 \times m_2 = -1$ (5.4).

Exam-style questions on straight lines

A typical calculator-allowed question sets a linear model in a real context and asks for interpretation as well as calculation.

Self-check

• To do:

  I can find the gradient of a line from two points, from a graph or from its equation, and state what its sign means.

• To do:

  I can convert between $y = mx + c$, $ax + by + d = 0$ and $y - y_1 = m(x - x_1)$, giving integer coefficients when asked.

• To do:

  I can find the $x$- and $y$-intercepts of a line and the area of the triangle it makes with the axes.

• To do:

  I can write the equations of horizontal and vertical lines and explain why a vertical line has no gradient.

• To do:

  I can decide whether two lines are parallel, perpendicular or neither, including when a parameter is involved.

• To do:

  I can find the point of intersection of two lines by substitution or elimination.

• To do:

  I can find the equation of the perpendicular bisector of a segment and use it to find a point equidistant from others.

• To do:

  I can build a linear model from a context and interpret its gradient and intercept with units.

[2]2 marks

Use the point $P(-1, 4)$ in the point–gradient form.

Multiply every term by $2$ to clear the fraction, then collect all terms on one side.

Check with $Q(3, -2)$: $3(3) + 2(-2) - 5 = 9 - 4 - 5 = 0$. ✓

so $L$ meets the $x$-axis at $(-4, 0)$.

On the $y$-axis $x = 0$:

so $L$ meets the $y$-axis at $(0, 3)$.

(b) The triangle has a right angle at the origin, a base of length $4$ along the $x$-axis and a height of $3$ along the $y$-axis. Lengths are positive, even though the intercept $-4$ is negative.

Multiply by $5$:

Method 2. A parallel line has the form $2x + 5y + d = 0$. Substitute $(4, -1)$:

Both methods give $2x + 5y - 3 = 0$.

Multiply by $2$ and rearrange:

Check: $6 + 2(1) - 8 = 0$, and the gradient of $x + 2y - 8 = 0$ is $-\frac{1}{2}$, with $2 \times \left(-\frac{1}{2}\right) = -1$. ✓

Gradients.

so the perpendicular gradient is $-\frac{3}{2}$.

Equation.

Multiply by $2$:

Check: $M(1, 3)$ gives $3 + 6 - 9 = 0$. ✓ The point $(3, 0)$ also lies on the bisector, and its distances to $A$ and $B$ are $\sqrt{5^2 + 1^2} = \sqrt{26}$ and $\sqrt{1^2 + 5^2} = \sqrt{26}$, equal as expected.

(b) The centre is equidistant from $A$, $B$ and $C$, so it lies on both perpendicular bisectors. Substitute $y = 0$ into $3x + 2y - 9 = 0$:

The centre is $(3, 0)$. The radius is its distance from $A$:

Check with $C$: $\sqrt{(4 - 3)^2 + (-5 - 0)^2} = \sqrt{26}$. ✓

The intercept is $32$, so

(b) Each rise of $1\,°\text{C}$ corresponds to a rise of $1.8\,°\text{F}$.

(c) Put $F = C$:

Both scales read $-40$ at the same temperature.

(b) The midpoint of $AB$ is

The perpendicular gradient is $-3$, since $\frac{1}{3} \times (-3) = -1$. Using $M$:

(c) $L$ has gradient $\frac{1}{3}$ and $y$-intercept $-6$, so $L$ is $y = \frac{1}{3}x - 6$. Substitute into $3x + y - 9 = 0$:

Then $y = \frac{1}{3} \times \frac{9}{2} - 6 = \frac{3}{2} - 6 = -\frac{9}{2}$, so $D = \left(\frac{9}{2}, -\frac{9}{2}\right)$.

(d) The perpendicular bisector is perpendicular to $AB$, and $L$ is parallel to $AB$, so the bisector meets $L$ at right angles at $D$. The shortest distance from $M$ to $L$ is therefore $MD$.

The word "hence" means the answer to (c) must be used, so the reason that $MD$ is perpendicular to $L$ must be stated.

(a) Parallel lines have equal gradients:

Check that the lines are not identical. For $k = 2$: $2x + 2y - 3 = 0$ and $3x + 3y + 5 = 0$, which are $x + y = 1.5$ and $x + y = -\frac{5}{3}$, distinct. For $k = -3$: $-3x + 2y - 3 = 0$ and $3x - 2y + 5 = 0$, which are $3x - 2y = -3$ and $3x - 2y = -5$, distinct. Both values give parallel lines.

(b) Perpendicular lines satisfy $m_1 m_2 = -1$:

Check: $m_1 = \frac{1}{5}$ and $m_2 = -\frac{3}{0.6} = -5$, and $\frac{1}{5} \times (-5) = -1$. ✓

(b) The temperature falls by $6.5\,°\text{C}$ for each kilometre gained in height.

(c) Solve $15 - 6.5h = 0$:

(d) At $h = 11$: $T = 15 - 6.5 \times 11 = 15 - 71.5 = -56.5\,°\text{C}$.

• y = f(x)
• L

[3]3 marks

[4]4 marks

Use this fact to show that $\tan\left(\dfrac{5\pi}{12}\right) = 2 + \sqrt{3}$.

[3]3 marks

Find the exact area of triangle OCD, where O is the origin.

[3]3 marks