A straight line is fixed by its gradient and one point on it. From that you can write its equation in any form the question asks for, decide whether two lines are parallel or perpendicular, find where lines cross and prove that three points lie on one line.
The gradient of the line through \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) is \(m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\). The line with gradient \(m\) through \((x_{1}, y_{1})\) is \(y - y_{1} = m(x - x_{1})\); rearrange it to \(y = mx + c\) or to \(ax + by + c = 0\) with integers as asked. Parallel lines have equal gradients, \(m_{1} = m_{2}\). Perpendicular lines have \(m_{1}m_{2} = -1\), so the perpendicular gradient is the negative reciprocal: \(\dfrac{2}{3}\) becomes \(-\dfrac{3}{2}\).
Fig. 1: y = ½x + 1 and y = ½x - 2 have the same gradient, so they are parallel. y = -2x + 6 meets y = ½x + 1 at right angles at (2, 2), because ½ × (-2) = -1
Section 01
Gradient and the three forms.
Find the gradient first: change in \(y\) divided by change in \(x\), taking both differences in the same order. Then put the gradient and any point on the line into \(y - y_{1} = m(x - x_{1})\). This point-gradient form is the quickest start; the question decides which form you finish in.
Form
Use it when
Gradient
\(y = mx + c\)
you want the gradient and the \(y\)-intercept \(c\) at a glance
\(m\)
\(y - y_{1} = m(x - x_{1})\)
you know the gradient and one point \((x_{1}, y_{1})\)
\(m\)
\(ax + by + c = 0\)
the question asks for integer coefficients
\(-\dfrac{a}{b}\): make \(y\) the subject to see it
Worked example Line through two points, integer form
Find the equation of the line through \(A(-2, 5)\) and \(B(4, 1)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.
Worked example Reading a line in the form ax + by + c = 0
Find the gradient of the line \(3x - 4y + 8 = 0\) and the coordinates of the points where it crosses the axes.
Make \(y\) the subject: \(4y = 3x + 8\), so \(y = \dfrac{3}{4}x + 2\).
Gradient \(\dfrac{3}{4}\); the line crosses the \(y\)-axis at \((0, 2)\).
On the \(x\)-axis \(y = 0\): \(3x + 8 = 0\), so \(x = -\dfrac{8}{3}\).
Answer: Gradient \(\frac{3}{4}\); crosses the axes at \((0, 2)\) and \(\left(-\frac{8}{3}, 0\right)\)
Examiner trap
Keep the two differences in the same order: \(\dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\), never \(\dfrac{y_{2} - y_{1}}{x_{1} - x_{2}}\), which flips the sign. And bracket negative coordinates: \(4 - (-2) = 6\), not \(2\). Read the gradient of \(ax + by + c = 0\) only after making \(y\) the subject; the coefficient of \(x\) is not the gradient.
Section 02
Parallel and perpendicular.
Parallel lines rise at the same rate, so they have the same gradient: \(m_{1} = m_{2}\). Two lines are perpendicular when the product of their gradients is \(-1\): \(m_{1}m_{2} = -1\), so \(m_{2} = -\dfrac{1}{m_{1}}\). Turn the fraction upside down AND change its sign (Fig. 1). The exception: a horizontal line \(y = k\) is perpendicular to a vertical line \(x = h\), whose gradient is undefined.
Worked example Through a point, parallel to a given line
Find the equation of the line through \((3, -1)\) that is parallel to \(2x + 5y = 7\). Give your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.
Gradient of \(2x + 5y = 7\): \(5y = -2x + 7\), so \(m = -\dfrac{2}{5}\).
A parallel line has the same gradient: \(y + 1 = -\dfrac{2}{5}(x - 3)\).
Multiply by 5: \(5y + 5 = -2x + 6\).
Rearrange: \(2x + 5y - 1 = 0\). It has the same \(2x + 5y\) as the given line; only the constant changes.
Answer: \(2x + 5y - 1 = 0\)
Worked example Through a point, perpendicular to a given line
Find the equation of the line through \((-4, 2)\) that is perpendicular to \(3x - 2y = 6\). Give your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.
Gradient of \(3x - 2y = 6\): \(2y = 3x - 6\), so \(m_{1} = \dfrac{3}{2}\).
Show that the lines \(y = 3x - 1\) and \(x + 3y = 9\) are perpendicular.
The first has gradient \(3\). For the second, \(3y = -x + 9\), so \(y = -\dfrac{1}{3}x + 3\), gradient \(-\dfrac{1}{3}\). The product is \(3 \times \left(-\dfrac{1}{3}\right) = -1\), so the lines are perpendicular.
Examiner trap
The perpendicular gradient needs BOTH steps: \(-\dfrac{2}{5}\) becomes \(\dfrac{5}{2}\). Writing \(-\dfrac{5}{2}\) (reciprocal only) or \(\dfrac{2}{5}\) (sign change only) gives a line that is not at right angles, and every later mark is lost.
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 \((1, -2)\) and \((5, 6)\).
Quick check+10 XP
What is the gradient of the line \(4x - 2y + 3 = 0\)?
Quick check+10 XP
A line has gradient \(-\dfrac{3}{4}\). What is the gradient of a line perpendicular to it?
Quick check+10 XP
Which line is parallel to \(y = 3x - 5\)?
Section 03
Crossing points, collinear points and unknowns.
Where two lines meet, the coordinates satisfy both equations, so solve them simultaneously; substitution is quickest when one equation already has \(y\) as the subject. Three points \(A\), \(B\), \(C\) are collinear (on one straight line) when the gradient of \(AB\) equals the gradient of \(BC\): the shared point \(B\) means the two segments cannot be separate parallel lines.
Worked example Point of intersection
Find the coordinates of the point where the lines \(2x + 3y = 12\) and \(y = x - 1\) meet.
Substitute \(y = x - 1\) into the first: \(2x + 3(x - 1) = 12\).
\(5x - 3 = 12\), so \(x = 3\).
\(y = 3 - 1 = 2\). Check in the first: \(6 + 6 = 12\).
Answer: \((3, 2)\)
Worked example Showing three points are collinear
Show that \(A(-1, 4)\), \(B(2, -2)\) and \(C(4, -6)\) are collinear, and find the equation of the line through them.
The lines \(kx + 2y = 5\) and \(3x - 6y = 1\) are perpendicular. Find the value of \(k\).
Gradients: \(y = -\dfrac{k}{2}x + \dfrac{5}{2}\) gives \(-\dfrac{k}{2}\); \(6y = 3x - 1\) gives \(\dfrac{1}{2}\). Then \(-\dfrac{k}{2} \times \dfrac{1}{2} = -1\), so \(-\dfrac{k}{4} = -1\) and \(k = 4\).
Examiner trap
Equal gradients alone do not prove collinearity: \(AB\) and \(CD\) with the same gradient may be two different parallel lines. Compare two gradients that SHARE a point (\(AB\) and \(BC\)), and say so in your conclusion.
Stage 2 · Exam
Exam-style questions
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Exam style+20 XP
Find the equation of the line through \((2, -3)\) that is perpendicular to \(5x - 2y = 4\), in the form \(ax + by + c = 0\) with integer coefficients.
Exam style+20 XP
Find the coordinates of the point of intersection of the lines \(y = 2x - 3\) and \(3x + 2y = 15\).
Exam style+20 XP
The points \(A(1, 2)\), \(B(3, k)\) and \(C(7, 14)\) are collinear. Find the value of \(k\).
Exam style+20 XP
The lines \((k + 1)x + 3y = 4\) and \(2x - y = 7\) are perpendicular. Find the value of \(k\).
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