Equations of straight lines

AS · P1 · 14 min

The equation of a line is a rule that every point on the line obeys and no other point does. Paper 1 expects you to find that rule from "sufficient information" (two points, or a point and a gradient), to move between the three standard forms, and to use lines to find intersections, perpendicular bisectors, feet of perpendiculars and areas. These skills appear in nearly every paper, either as a standalone question or as the first part of a circle or calculus question.

What an equation of a line means

A point $(x, y)$ lies on a line exactly when its coordinates satisfy the line's equation. So:

to check whether a point is on a line, substitute it in;
to find where two lines meet, find the point that satisfies both equations at once;
to find where a line meets an axis, put $y = 0$ (for the $x$ -axis) or $x = 0$ (for the $y$ -axis).

Every line that is not vertical has a gradient $m$ (see Length, gradient and midpoint), and its equation can be built from the gradient and one point.

The three forms

Gradient and one point

Suppose a line has gradient $m$ and passes through the fixed point $(x_1, y_1)$ . Take any other point $(x, y)$ on the line. The gradient from $(x_1, y_1)$ to $(x, y)$ must be $m$ :

\frac{y - y_1}{x - x_1} = m \quad\Rightarrow\quad y - y_1 = m(x - x_1)

This is the most useful form, because exam questions nearly always give you a point and a way to find the gradient.

Gradient and intercept

Choosing the point $(0, c)$ , where the line crosses the $y$ -axis, gives $y - c = m(x - 0)$ , that is $y = mx + c$ . Here $c$ is the $y$ -intercept. This form is best for reading off the gradient and for sketching.

The general form

Multiplying out and collecting everything on one side gives $ax + by + c = 0$ . Questions often ask for this form "where $a$ , $b$ and $c$ are integers", which simply means clear the fractions.

Key result

Form	When to use it
$y - y_1 = m(x - x_1)$	you know a point and the gradient (the usual starting point)
$y = mx + c$	you want to read off the gradient or $y$ -intercept, or sketch
$ax + by + c = 0$	the question asks for integer coefficients

To read the gradient from $ax + by + c = 0$ , rearrange to $y = -\dfrac{a}{b}x - \dfrac{c}{b}$ , so the gradient is $-\dfrac{a}{b}$ .

Vertical lines have equations $x = k$ ; horizontal lines have equations $y = k$ .

Finding the equation of a line

Find the gradient $m$ : from two points, or from a parallel line (same $m$ ), or from a perpendicular line ( $-\tfrac{1}{m}$ ).
Choose a point on the line.
Substitute into $y - y_1 = m(x - x_1)$ .
Rearrange into the form asked for. If integer coefficients are wanted, multiply through by the denominator.
Check: substitute your known point(s) back into the final equation.

y = -2x + 3 (-1, 5) (3, -3) (0, 3)

The line through $(-1, 5)$ and $(3, -3)$ has gradient $-2$ and $y$ -intercept $3$ : $y = -2x + 3$ , or $2x + y - 3 = 0$ .

Perpendicular bisectors

The perpendicular bisector of $AB$ is the line that passes through the midpoint of $AB$ and is perpendicular to $AB$ . Every point on it is the same distance from $A$ as from $B$ . That property makes it essential for circles: the centre of any circle through $A$ and $B$ lies on the perpendicular bisector of $AB$ (see Circles).

Perpendicular bisector of AB

Find the midpoint $M$ of $AB$ .
Find the gradient of $AB$ , then the perpendicular gradient $-\tfrac{1}{m}$ .
Use $y - y_M = -\tfrac{1}{m}(x - x_M)$ .

Intersections of two lines

Two lines that are not parallel meet at exactly one point. Solve their equations simultaneously, by substitution if one is already in the form $y = \ldots$ , and by elimination otherwise. Parallel lines (equal gradients, different intercepts) never meet: the simultaneous equations have no solution.

The foot of a perpendicular and the distance to a line

The shortest distance from a point $C$ to a line $l$ is measured along the perpendicular from $C$ to $l$ . The point $N$ where this perpendicular meets $l$ is called the foot of the perpendicular.

Foot of the perpendicular and shortest distance

Find the gradient of $l$ , then the perpendicular gradient.
Write the equation of the line through $C$ with the perpendicular gradient.
Solve simultaneously with $l$ to find $N$ .
The shortest distance from $C$ to $l$ is the length $CN$ .

The same construction gives the reflection of $C$ in $l$ : the image $C'$ is on the other side, with $N$ as the midpoint of $CC'$ .

y = 2x + 1 y = -0.5x + 3.5 (7, 0) (1, 3) (-5, 6)

In the figure, $C(7, 0)$ , the line $y = 2x + 1$ , the foot $N(1, 3)$ and the reflection $C'(-5, 6)$ .

Tip

There is a formula for the perpendicular distance from a point to a line, but it is not on the 9709 syllabus and is not in the formula list. The method above always works and shows the examiner every step.

Worked examples

Through two points

Find the equation of the line through $A(-1, 5)$ and $B(3, -3)$ , giving your answer in the form $ax + by + c = 0$ .

Solution

m = \frac{-3 - 5}{3 - (-1)} = \frac{-8}{4} = -2

Using $A$ :

y - 5 = -2(x + 1) \quad\Rightarrow\quad y - 5 = -2x - 2 \quad\Rightarrow\quad 2x + y - 3 = 0

Check with $B$ : $2(3) + (-3) - 3 = 0$ . Correct.

Perpendicular to a given line

Find the equation of the line through $(4, -1)$ that is perpendicular to $3x - 2y = 6$ .

Solution

Rearrange the given line: $2y = 3x - 6$ , so $y = \tfrac{3}{2}x - 3$ , gradient $\tfrac{3}{2}$ . The perpendicular gradient is $-\tfrac{2}{3}$ .

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

Multiply by $3$ : $3y + 3 = -2x + 8$ , so $2x + 3y = 5$ .

A perpendicular bisector

The points $P$ and $Q$ are $(2, 1)$ and $(8, 5)$ . Find the equation of the perpendicular bisector of $PQ$ .

Solution

Midpoint $M = (5, 3)$ . Gradient of $PQ = \dfrac{5 - 1}{8 - 2} = \dfrac{2}{3}$ , so the perpendicular gradient is $-\dfrac{3}{2}$ .

y - 3 = -\tfrac{3}{2}(x - 5) \quad\Rightarrow\quad 2y - 6 = -3x + 15 \quad\Rightarrow\quad 3x + 2y = 21

Check: $M$ gives $15 + 6 = 21$ .

Foot of a perpendicular, distance and reflection

The line $l$ has equation $y = 2x + 1$ and the point $C$ is $(7, 0)$ .

(a) Find the coordinates of the foot of the perpendicular from $C$ to $l$ .

(b) Find the shortest distance from $C$ to $l$ .

Solution

(a) The perpendicular gradient is $-\tfrac{1}{2}$ . The line through $C$ : $y = -\tfrac{1}{2}(x - 7)$ . At the intersection with $l$ :

2x + 1 = -\tfrac{1}{2}x + \tfrac{7}{2} \quad\Rightarrow\quad 4x + 2 = -x + 7 \quad\Rightarrow\quad x = 1

and $y = 2(1) + 1 = 3$ . The foot is $N(1, 3)$ .

(b) $CN = \sqrt{(7 - 1)^2 + (0 - 3)^2} = \sqrt{45} = 3\sqrt{5}$ .

A right angle at a vertex and an area

The points $A$ and $B$ are $(1, 3)$ and $(5, 11)$ . The line through $A$ perpendicular to $AB$ meets the $x$ -axis at $C$ .

(a) Find the equation of $AC$ and the coordinates of $C$ .

(b) Find the area of triangle $ABC$ .

Solution

(a) $m_{AB} = \dfrac{11 - 3}{5 - 1} = 2$ , so $m_{AC} = -\dfrac{1}{2}$ :

y - 3 = -\tfrac{1}{2}(x - 1) \quad\Rightarrow\quad x + 2y = 7

At $C$ , $y = 0$ , so $x = 7$ : $C = (7, 0)$ .

(b) The angle at $A$ is $90^\circ$ , so $AB$ and $AC$ are the base and height:

AB = \sqrt{4^2 + 8^2} = \sqrt{80}, \qquad AC = \sqrt{6^2 + 3^2} = \sqrt{45}

\text{Area} = \tfrac{1}{2}\sqrt{80}\sqrt{45} = \tfrac{1}{2}\sqrt{3600} = 30

A line through a fixed point with unknown gradient

The line $l$ passes through $(2, 3)$ and has negative gradient $m$ . It meets the $x$ -axis at $P$ and the $y$ -axis at $Q$ , and $O$ is the origin.

(a) Find the coordinates of $P$ and $Q$ in terms of $m$ .

(b) Given that the area of triangle $OPQ$ is $12$ , find $m$ and the equation of $l$ .

Solution

(a) The line is $y - 3 = m(x - 2)$ .

At $Q$ , $x = 0$ : $y = 3 - 2m$ , so $Q = (0,\ 3 - 2m)$ .

At $P$ , $y = 0$ : $-3 = m(x - 2)$ , so $x = 2 - \dfrac{3}{m}$ and $P = \left(2 - \dfrac{3}{m},\ 0\right)$ .

Since $m < 0$ , both $3 - 2m$ and $2 - \tfrac{3}{m}$ are positive, so $P$ and $Q$ are on the positive axes.

(b)

\tfrac{1}{2}\left(2 - \frac{3}{m}\right)(3 - 2m) = 12

Multiply by $2m$ (remembering $m \neq 0$ ):

\begin{aligned} (2m - 3)(3 - 2m) &= 24m \\ -(2m - 3)^2 &= 24m \\ -4m^2 + 12m - 9 &= 24m \\ 4m^2 + 12m + 9 &= 0 \\ (2m + 3)^2 &= 0 \end{aligned}

So $m = -\tfrac{3}{2}$ (a repeated root, so this is the only such line). Then $P = (4, 0)$ , $Q = (0, 6)$ , and $l$ is

y - 3 = -\tfrac{3}{2}(x - 2) \quad\Rightarrow\quad 3x + 2y = 12

Check: area $= \tfrac{1}{2} \times 4 \times 6 = 12$ . The repeated root means $12$ is the smallest area any such triangle can have.

Watch out

Using the gradient of the wrong line. In "perpendicular to $3x - 2y = 6$ ", the gradient of the given line is $\tfrac{3}{2}$ , not $3$ or $-2$ . Rearrange to $y = mx + c$ before reading it off.

Dropping a sign when multiplying out. $y - 5 = -2(x + 1)$ gives $y - 5 = -2x - 2$ . Write the bracket expansion as a separate line.

Leaving fractions when integers are asked for. " $y = -\tfrac{2}{3}x + \tfrac{5}{3}$ " is correct but not in the form $ax + by = c$ with integer coefficients; multiply by $3$ .

Confusing the perpendicular bisector with any perpendicular line. It must pass through the midpoint.

Reading the shortest distance as the distance to an intercept. The shortest distance to a line is along the perpendicular, not to where the line crosses an axis.

Exam tip

"Sufficient information". If a question gives you a point and a direction (parallel, perpendicular, through another point), you have enough. Start with $y - y_1 = m(x - x_1)$ .
Form of the answer. Follow the requested form exactly. "In the form $y = mx + c$ " and "in the form $ax + by + c = 0$ " are different, and the final mark is for the right form.
Exact coordinates. Keep fractions exact. Intersections and feet of perpendiculars are often reused in later parts, and rounded values lose accuracy marks there.
Areas. If a triangle has a right angle or a side along an axis, use $\tfrac{1}{2} \times$ base $\times$ height. For other triangles, see the methods in Proving properties of shapes.
Check one point. Substituting a known point into your final equation takes ten seconds and catches most slips.

Summary

$y - y_1 = m(x - x_1)$ : line with gradient $m$ through $(x_1, y_1)$ . Use it as the default.
$y = mx + c$ : gradient $m$ , $y$ -intercept $c$ . $ax + by + c = 0$ has gradient $-\tfrac{a}{b}$ .
Vertical lines are $x = k$ ; horizontal lines are $y = k$ .
Parallel lines share a gradient; perpendicular gradients multiply to $-1$ .
Perpendicular bisector: through the midpoint, perpendicular to the segment.
Intersection: solve simultaneously. Axis intercepts: put $x = 0$ or $y = 0$ .
Shortest distance to a line: find the foot of the perpendicular, then the length. The reflection has the foot as its midpoint.

Practice questions

Question

Find the equation of the line with gradient $4$ through $(3, -2)$ , in the form $y = mx + c$ .
Find the equation of the line through $(-2, 5)$ and $(4, 2)$ , in the form $ax + by + c = 0$ with integer coefficients.
Find the equation of the line through $(1, 4)$ parallel to $2x + 5y = 7$ .
Find the equation of the perpendicular bisector of the line joining $(-3, 2)$ and $(5, 6)$ .
Find the point of intersection of $3x - 2y = 4$ and $x + 4y = 13$ .
The line $5x + 2y = 20$ meets the $x$ -axis at $A$ and the $y$ -axis at $B$ . Find the exact length of $AB$ and the area of triangle $OAB$ , where $O$ is the origin.
Find the foot of the perpendicular from $(5, 6)$ to the line $x + 2y = 7$ . Hence find the shortest distance from $(5, 6)$ to the line, and the reflection of $(5, 6)$ in the line.
The lines $y = 3x - 2$ and $x + 2y = 10$ meet at $P$ . Find the equation of the line through $P$ perpendicular to $x + 2y = 10$ .
The points $A$ and $B$ are $(-1, 1)$ and $(3, 9)$ . The line through $B$ perpendicular to $AB$ meets the $x$ -axis at $C$ . Find the coordinates of $C$ and the area of triangle $ABC$ .
The line $l_1$ has equation $x + 2y = 9$ . The line $l_2$ passes through $A(0, -3)$ and is perpendicular to $l_1$ . (a) Find the equation of $l_2$ and the point $B$ where $l_1$ and $l_2$ meet. (b) Find the reflection of $A$ in $l_1$ . (c) Find the shortest distance from $A$ to $l_1$ , in exact form.

Answers

$y + 2 = 4(x - 3)$ , so $y = 4x - 14$ .
$m = \dfrac{2 - 5}{4 - (-2)} = -\dfrac{1}{2}$ . $y - 5 = -\tfrac{1}{2}(x + 2)$ ; multiply by $2$ : $2y - 10 = -x - 2$ , so $x + 2y - 8 = 0$ .
$2x + 5y = 7$ has gradient $-\tfrac{2}{5}$ . $y - 4 = -\tfrac{2}{5}(x - 1)$ , so $5y - 20 = -2x + 2$ , giving $2x + 5y = 22$ . (A parallel line keeps the left-hand side $2x + 5y$ ; only the constant changes.)
Midpoint $(1, 4)$ ; gradient of segment $\tfrac{4}{8} = \tfrac{1}{2}$ , so perpendicular gradient $-2$ . $y - 4 = -2(x - 1)$ , so $2x + y = 6$ .
From the second equation $x = 13 - 4y$ . Then $3(13 - 4y) - 2y = 4$ , so $39 - 14y = 4$ and $y = \tfrac{5}{2}$ , $x = 3$ . The point is $\left(3, \tfrac{5}{2}\right)$ .
$A = (4, 0)$ and $B = (0, 10)$ . $AB = \sqrt{16 + 100} = \sqrt{116} = 2\sqrt{29}$ . Area $= \tfrac{1}{2} \times 4 \times 10 = 20$ .
The line has gradient $-\tfrac{1}{2}$ , so the perpendicular has gradient $2$ : $y - 6 = 2(x - 5)$ , i.e. $y = 2x - 4$ . Substituting: $x + 2(2x - 4) = 7$ , so $5x = 15$ , $x = 3$ , $y = 2$ . Foot $(3, 2)$ . Distance $\sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}$ . Reflection: step from $(5, 6)$ to $(3, 2)$ is $(-2, -4)$ , so the image is $(1, -2)$ .
$x + 2(3x - 2) = 10$ , so $7x = 14$ , $x = 2$ , $y = 4$ : $P = (2, 4)$ . The perpendicular gradient to $-\tfrac{1}{2}$ is $2$ : $y - 4 = 2(x - 2)$ , so $y = 2x$ .
$m_{AB} = \tfrac{8}{4} = 2$ , so the perpendicular gradient is $-\tfrac{1}{2}$ : $y - 9 = -\tfrac{1}{2}(x - 3)$ . At $y = 0$ : $x - 3 = 18$ , so $C = (21, 0)$ . $AB = \sqrt{80}$ , $BC = \sqrt{18^2 + 9^2} = \sqrt{405}$ . The angle at $B$ is $90^\circ$ , so area $= \tfrac{1}{2}\sqrt{80 \times 405} = \tfrac{1}{2}\sqrt{32400} = \tfrac{1}{2}(180) = 90$ .
(a) $l_1$ has gradient $-\tfrac{1}{2}$ , so $l_2$ has gradient $2$ and equation $y = 2x - 3$ . Then $x + 2(2x - 3) = 9$ , so $5x = 15$ , $x = 3$ , $y = 3$ : $B = (3, 3)$ . (b) $B$ is the midpoint of $A$ and its image. Step $A \to B$ is $(3, 6)$ , so the image is $(6, 9)$ . (c) $AB = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}$ .

What an equation of a line means

The three forms

Gradient and one point

Gradient and intercept

The general form

Perpendicular bisectors

Intersections of two lines

The foot of a perpendicular and the distance to a line

Worked examples

Practice questions

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