A-Level Maths 9709 / Topic 1.3 / AS & A Level

Two facts fix a line.

A gradient and a point, or two points, are all you need to write the equation of a line. Use the three standard forms, the rule for perpendicular gradients, and distances and midpoints to solve geometry problems with algebra.

Full-screen labStraight lines lab → Topic 1.3All coordinate geometry lessons →

The key idea

The line through \((x_{1}, y_{1})\) with gradient \(m\) is \(y - y_{1} = m(x - x_{1})\). Parallel lines have equal gradients; perpendicular lines have \(m_{1}m_{2} = -1\), so a perpendicular gradient is the negative reciprocal \(-\dfrac{1}{m}\).

Fig. 1: the perpendicular bisector of A(1, 7) and B(5, -1) passes through the midpoint M(3, 3) with gradient ½, because AB has gradient -2

Section 01

Three forms of one equation.

In the lab, drag the two points and watch the gradient and the equation of the line update as you go. Use whichever form suits the information you have, and give the final answer in the form the question asks for.

Gradient through \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\): \(m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\).
\(y = mx + c\): gradient \(m\), \(y\)-intercept \(c\). Best for reading a line.
\(y - y_{1} = m(x - x_{1})\): best for building a line from a point and a gradient.
\(ax + by + c = 0\): often requested with integer \(a\), \(b\), \(c\). Rearrange to \(y = \ldots\) to read its gradient \(-\dfrac{a}{b}\).

Worked example Through two points

Find the equation of the line through \(A(-2, 5)\) and \(B(4, -1)\), in the form \(ax + by + c = 0\).

\(m = \dfrac{-1 - 5}{4 - (-2)} = \dfrac{-6}{6} = -1\).
Using \(A\): \(y - 5 = -1(x + 2) \Rightarrow y = -x + 3\).
Check with \(B\): \(-4 + 3 = -1\), as required. Rearrange so everything is on one side.

Answer: \(x + y - 3 = 0\)

Worked example Point and fractional gradient

Find the equation of the line through \((3, -2)\) with gradient \(\dfrac{2}{3}\), in the form \(ax + by + c = 0\) with integers \(a\), \(b\), \(c\).

\(y + 2 = \dfrac{2}{3}(x - 3)\).
Multiply by 3 to clear the fraction: \(3y + 6 = 2x - 6\).
Collect on one side.

Answer: \(2x - 3y - 12 = 0\)

Worked example Reading a line

Find the gradient of \(3x + 4y = 24\) and the points where it crosses the axes.

\(4y = -3x + 24 \Rightarrow y = -\dfrac{3}{4}x + 6\), so \(m = -\dfrac{3}{4}\).
\(x = 0\): \(y = 6\). \(y = 0\): \(3x = 24\), \(x = 8\).

Answer: Gradient \(-\dfrac{3}{4}\); crosses at \((0, 6)\) and \((8, 0)\)

Examiner trap

The gradient of \(3x + 4y = 24\) is not 3 (nor \(\frac{4}{3}\)). Always rearrange to \(y = mx + c\) before reading \(m\).

Section 02

Parallel and perpendicular.

In the lab, drag the points to turn the line and watch its perpendicular turn with it: the product of their gradients stays at \(-1\). The perpendicular bisector of \(AB\) (Fig. 1) is the classic exam use: it passes through the midpoint at right angles to \(AB\).

Worked example Perpendicular bisector

Find the equation of the perpendicular bisector of \(A(1, 7)\) and \(B(5, -1)\) (Fig. 1).

Midpoint \(M = \left(\dfrac{1 + 5}{2},\ \dfrac{7 + (-1)}{2}\right) = (3, 3)\).
Gradient of \(AB = \dfrac{-1 - 7}{5 - 1} = -2\), so the perpendicular gradient is \(\dfrac{1}{2}\).
\(y - 3 = \dfrac{1}{2}(x - 3) \Rightarrow 2y - 6 = x - 3\).

Answer: \(x - 2y + 3 = 0\)

Worked example Parallel line

Find the equation of the line through \((2, 1)\) parallel to \(4x - 2y = 7\).

\(4x - 2y = 7 \Rightarrow y = 2x - \dfrac{7}{2}\), so the gradient is 2.
Parallel, so the same gradient: \(y - 1 = 2(x - 2)\).

Answer: \(y = 2x - 3\)

Your turn

Find the equation of the line through \((2, -3)\) perpendicular to \(x + 4y = 1\).

Examiner trap

The perpendicular gradient is the NEGATIVE reciprocal. For \(m = -2\) it is \(+\frac{1}{2}\); \(-\frac{1}{2}\) (reciprocal only) and \(2\) (sign change only) are both wrong.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Find the gradient of the line through \((2, 3)\) and \((6, 11)\).

Quick check+10 XP

Find the gradient of a line perpendicular to \(y = 4x - 1\).

Quick check+10 XP

Find the gradient of \(2x + 5y = 10\).

Quick check+10 XP

Find the midpoint of \((-3, 4)\) and \((7, -2)\).

Section 03

Lengths, midpoints and intersections.

Most coordinate geometry questions combine several small tools. Draw a quick sketch, label every point, and work one step at a time.

Length: \(AB = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}\).
Midpoint: \(\left(\dfrac{x_{1} + x_{2}}{2},\ \dfrac{y_{1} + y_{2}}{2}\right)\).
Intersection: solve the two equations simultaneously.

Worked example Intersection

Find the point of intersection of \(2x + y = 7\) and \(x - 3y = 7\).

From the second equation, \(x = 7 + 3y\).
Substitute: \(2(7 + 3y) + y = 7 \Rightarrow 14 + 7y = 7 \Rightarrow y = -1\).
\(x = 7 + 3(-1) = 4\).

Answer: \((4,\ -1)\)

Worked example A right-angled triangle

\(A(0, 2)\) and \(B(6, 5)\). The line through \(B\) perpendicular to \(AB\) meets the \(x\)-axis at \(C\). Find \(C\) and the area of triangle \(ABC\).

Gradient \(AB = \dfrac{3}{6} = \dfrac{1}{2}\), so the perpendicular gradient is \(-2\): \(y - 5 = -2(x - 6) \Rightarrow y = -2x + 17\).
\(y = 0\): \(x = \dfrac{17}{2}\), so \(C = \left(\dfrac{17}{2}, 0\right)\).
\(AB = \sqrt{6^{2} + 3^{2}} = \sqrt{45} = 3\sqrt{5}\); \(BC = \sqrt{2.5^{2} + 5^{2}} = \sqrt{31.25} = \dfrac{5\sqrt{5}}{2}\).
The right angle is at \(B\), so area \(= \dfrac{1}{2} \times 3\sqrt{5} \times \dfrac{5\sqrt{5}}{2} = \dfrac{75}{4}\).

Answer: \(C\left(\dfrac{17}{2}, 0\right)\); area \(= \dfrac{75}{4} = 18.75\) square units

Examiner trap

In the length formula, subtract the coordinates before squaring: from \((-1, 3)\) to \((5, -5)\) the changes are 6 and \(-8\), giving \(\sqrt{36 + 64} = 10\). Adding the coordinates (4 and \(-2\)) is a common slip.

Stage 2 · Exam

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Find the equation of the line through \((-1, 4)\) perpendicular to \(3x - 2y + 5 = 0\).

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\(A(2, -1)\) and \(B(8, 7)\). Find the equation of the perpendicular bisector of \(AB\).

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Find the distance between \((-1, 3)\) and \((5, -5)\).

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The lines \(y = 2x - 1\) and \(3x + 2y = 12\) meet at P. Find the area of the triangle formed by the two lines and the \(y\)-axis.

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Six original Cambridge-style questions

Reading a gradient, a line through two points, a perpendicular bisector, a right-angled triangle, a proof that a triangle is right-angled and isosceles, and the closest approach of a ship to a lighthouse.