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Equations of straight lines in the forms \(y = mx + c\), \(y - y_1 = m(x - x_1)\) and \(ax + by + c = 0\), gradients, and the conditions for two lines to be parallel or perpendicular. It includes straight-line models in context. Line questions often combine with circles, intersection points and areas of triangles.
Key facts
\(m = \dfrac{y_2 - y_1}{x_2 - x_1}\)
\(y - y_1 = m(x - x_1)\)
\(y = mx + c\) and \(ax + by + c = 0\)
Gradient of \(ax + by + c = 0\) is \(-\dfrac{a}{b}\)
\(y - y_1 = m(x - x_1)\): the line through \((x_1, y_1)\) with gradient \(m\). This is usually the quickest form to write down.
\(ax + by + c = 0\): exam questions often ask for integers \(a\), \(b\) and \(c\). Its gradient is \(-\dfrac{a}{b}\)
Example: the line through \((2, -1)\) with gradient \(\dfrac{3}{4}\) is \(y + 1 = \dfrac{3}{4}(x - 2)\), so \(4y + 4 = 3x - 6\), i.e. \(3x - 4y - 10 = 0\)
Parallel and perpendicular lines
diagram
Parallel lines have equal gradients.
Perpendicular lines: \(m_1 \times m_2 = -1\), so the perpendicular gradient is \(-\dfrac{1}{m}\) (flip and change the sign). E.g. \(\dfrac{2}{3}\) becomes \(-\dfrac{3}{2}\)
The perpendicular bisector of \(AB\) passes through the midpoint of \(AB\) and is perpendicular to \(AB\)
The perpendicular bisector goes through the midpoint M(3, 3) and has gradient −1/m: here ½, because AB has gradient −2.
Two lines meet where their equations are both true: solve them simultaneously.
A line meets the \(x\)-axis where \(y = 0\) and the \(y\)-axis where \(x = 0\)
For the area of a triangle, sketch it and look for a side along an axis or a right angle, then use \(\dfrac{1}{2} \times\) base \(\times\) height.
Straight-line models
In a linear model the gradient is the rate of change (e.g. cm per kg) and the intercept is the value when the variable is 0 (e.g. a fixed charge).
A linear model is only valid over a sensible range of values; predictions far outside the data are unreliable.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Find the equation of a line through two points
4 marksGrade D
Find the gradient.
Use \(y - y_1 = m(x - x_1)\) with one of the points.
Rearrange into the form asked for, with integer coefficients.
Example. The line \(l\) passes through the points \(A(-3, 2)\) and \(B(5, 8)\). Find an equation for \(l\) in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.
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\(m = \dfrac{8 - 2}{5 - (-3)}\)
(M1)
\(m = \dfrac{3}{4}\)
(A1)
\(y - 2 = \dfrac{3}{4}(x + 3)\)
(M1)
\(3x - 4y + 17 = 0\)
(A1)
Find a perpendicular bisector
5 marksGrade B
Find the midpoint of the segment.
Find the gradient of the segment, then the perpendicular gradient.
Use \(y - y_1 = m(x - x_1)\) through the midpoint.
Example. The points \(A\) and \(B\) have coordinates \((1, 7)\) and \((5, -1)\). Find an equation of the perpendicular bisector of \(AB\), giving your answer in the form \(ax + by + c = 0\)
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Midpoint \((3, 3)\)
(B1)
Gradient of \(AB = \dfrac{-1 - 7}{5 - 1} = -2\)
(M1)
Perpendicular gradient \(\dfrac{1}{2}\)
(A1)
\(y - 3 = \dfrac{1}{2}(x - 3)\)
(M1)
\(x - 2y + 3 = 0\)
(A1)
Intersections and the area of a triangle
7 marksGrade A
Find the equation of the missing line.
Solve simultaneously to find where the lines meet.
Find the axis intercepts.
Sketch the triangle and use \(\dfrac{1}{2} \times\) base \(\times\) height.
Example. The line \(l_1\) has equation \(y = 2x - 2\). The line \(l_2\) is perpendicular to \(l_1\) and passes through the point \(B(11, 0)\). (a) Find an equation for \(l_2\). (b) Find the coordinates of the point \(P\) where \(l_1\) and \(l_2\) meet. (c) The line \(l_1\) crosses the \(x\)-axis at \(A\). Find the area of triangle \(APB\)
Find the gradient from two data points and form the equation.
Interpret the gradient and intercept in context.
Comment on when the model might not be valid.
Example. The length, \(L\) cm, of a spring is modelled as a linear function of the mass, \(m\) kg, hanging from it. When \(m = 2\), \(L = 15\) and when \(m = 5\),\(L = 21\). (a) Find an equation for \(L\) in terms of \(m\). (b) Interpret the gradient in context. (c) Give a reason why the model may not be valid when \(m = 100\)
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(a) Gradient \(\dfrac{21 - 15}{5 - 2} = 2\)
(M1)
\(L = 2m + 11\)
(A1)
(b) The spring stretches by 2 cm for each extra kg.
(B1)
(c) e.g. 100 kg is far outside the data, and the spring may pass its elastic limit or break.
(B1)
Don’t lose marks
Subtracting the coordinates in different orders on the top and bottom of the gradient fraction.
Using \(-m\) or \(\dfrac{1}{m}\) for a perpendicular gradient instead of \(-\dfrac{1}{m}\)
Leaving fractions when the question asks for integers \(a\), \(b\) and \(c\)
Writing \(ax + by + c\) without '\(= 0\)', so it is not an equation.
More tips
Memory tricks
Perpendicular gradient: flip and change the sign, \(\dfrac{a}{b} \to -\dfrac{b}{a}\)
Sketch the points roughly: it catches sign errors and makes areas obvious.
Check your line by substituting the given point(s) into your final equation.
If no form is specified, any correct equation scores; if a form is specified (e.g. integers \(a\),\(b\),\(c\)), you must give it.
Show the gradient calculation clearly; it is usually the first method mark.
In multi-part questions, a quick sketch of the lines and points shows which lengths and heights to use.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Find the gradient between two points\(m = \dfrac{y_2 - y_1}{x_2 - x_1}\): the change in \(y\) divided by the change in \(x\)
Grade E
Find a line from a point and gradientUse \(y - y_1 = m(x - x_1)\), then rearrange if a particular form is asked for.
Grade D
Give a line's equation with integer coefficientsClear any fractions and collect all the terms on one side, with \(a\), \(b\) and \(c\) integers.
Grade D
Use the parallel and perpendicular gradient rulesParallel lines have equal gradients; perpendicular gradients multiply to give \(-1\)
Grade C
Find midpoints, lengths and intersectionsUse the midpoint and distance formulae, and solve the equations of two lines simultaneously.
Grade C
Use straight-line models in contextInterpret the gradient as a rate of change and the intercept as a starting or fixed value.
Grade B
Find the perpendicular bisector of a segmentUse the midpoint of the segment and the negative reciprocal of its gradient.
Grade A
Solve multi-step problems such as triangle areasFind vertices from intercepts and intersections, then use \(\dfrac{1}{2} \times\) base \(\times\) height.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
The points \(A(2, 6)\) and \(B(4, 7)\) lie on the straight line \(l\).
(a) Find an equation for \(l\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.[3]
(b) Find the exact length of \(AB\).[2]
Show the answer and mark scheme
(a)Answer: \(x - 2y + 10 = 0\)
M1 for an attempt at the gradient, e.g. \(\frac{7 - 6}{4 - 2} = \frac{1}{2}\)
M1 for \(y - 6 = \frac{1}{2}(x - 2)\) or equivalent, using their gradient with either point (or \(y = mx + c\) with an attempt to find \(c\))
A1 for \(x - 2y + 10 = 0\) or any integer multiple
Worked solution: Gradient of \(l\) \(= \frac{7 - 6}{4 - 2} = \frac{1}{2}\) \(y - 6 = \frac{1}{2}(x - 2)\) Multiplying out and collecting terms on one side: \(x - 2y + 10 = 0\)
The points \(A(-3, 5)\) and \(B(-5, 3)\) are the ends of the line segment \(AB\).
(a) Find an equation of the perpendicular bisector of \(AB\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.[4]
(b) The perpendicular bisector of \(AB\) meets the \(x\)-axis at the point \(C\). Find the exact area of triangle \(ABC\).[3]
Show the answer and mark scheme
(a)Answer: \(x + y = 0\)
B1 for the midpoint of \(AB\), \((-4, 4)\)
M1 for the gradient of \(AB\) \(= 1\) and use of \(m_1 m_2 = -1\) to obtain \(-1\)
dM1 for \(y - 4 = -(x + 4)\) or equivalent, using their midpoint and their perpendicular gradient
A1 for \(x + y = 0\) or any integer multiple
Worked solution: Midpoint of \(AB\) \(= \left(\frac{-3 + (-5)}{2}, \frac{5 + 3}{2}\right) = (-4, 4)\) Gradient of \(AB\) \(= \frac{3 - 5}{-5 - (-3)} = 1\), so the perpendicular gradient is \(-1\). \(y - 4 = -(x + 4)\) which rearranges to \(x + y = 0\).
(b)Answer: \(8\)
M1 for substituting \(y = 0\) into their perpendicular bisector to find \(C(0, 0)\)
M1 for a complete method for the area, e.g. \(\frac{1}{2} \times AB \times MC\) where \(M\) is the midpoint of \(AB\), or the shoelace formula
A1 for \(8\) cao
Worked solution: At \(C\), \(y = 0\): \(x = 0\) so \(C(0, 0)\). \(AB = \sqrt{(-2)^2 + (-2)^2} = 2\sqrt{2}\) and, with \(M(-4, 4)\), \(MC = \sqrt{4^2 + 4^2} = 4\sqrt{2}\). \(C\) is on the perpendicular bisector, so \(CM\) is perpendicular to \(AB\): \(\text{Area} = \frac{1}{2} \times 2\sqrt{2} \times 4\sqrt{2} = 8\)
Question 3Hard7 marks
The points \(A\) and \(B\) have coordinates \((-2, 4)\) and \((6, -2)\) respectively. The line \(l\) has equation \(y = 3x - 5\). The point \(C\) lies on \(l\) such that the area of triangle \(ABC\) is \(45\).
Find the two possible sets of coordinates of \(C\).[7]
Show the answer and mark scheme
Answer: \(C(5, 10)\) or \(C(-1, -8)\)
M1 for a general point on \(l\), e.g. \(C\left(t, 3t - 5\right)\)
M1 for a correct method for the area of triangle \(ABC\) in terms of \(t\), e.g. the shoelace formula or \(\frac{1}{2} \times \text{base} \times \text{height}\)
A1 for area \(= \frac{1}{2}\left|30t - 60\right|\) oe
M1 for setting their area equal to \(45\) and considering both signs (or squaring)
A1 for \(t = 5\) and \(t = -1\)
M1 for substituting their values of \(t\) into the equation of \(l\)
A1 for \((5, 10)\) and \((-1, -8)\)
Worked solution: Let \(C = \left(t, 3t - 5\right)\). Area \(= \frac{1}{2}\left|(x_B - x_A)(y_C - y_A) - (x_C - x_A)(y_B - y_A)\right|\) \(= \frac{1}{2}\left|8\left(3t - 9\right) - (-6)\left(t + 2\right)\right| = \frac{1}{2}\left|30t - 60\right|\) Setting this equal to \(45\): \(30t - 60 = \pm 90\), so \(t = 5\) or \(t = -1\). Hence \(C(5, 10)\) or \(C(-1, -8)\).