Equation of a straight lineEdexcel International A Level Maths: Revision notes

Section 1

Gradient and the form y = mx + c

The gradient of the line through $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}.$ A line with gradient $m$ and $y$ -intercept $c$ has equation $y=mx+c$ . A positive gradient rises left to right, a negative one falls, and a horizontal line has $m=0$ . A vertical line has no gradient and is written $x=k$ . Example: through $A(2,5)$ and $B(6,13)$ , $m=\frac{13-5}{6-2}=2$ .

Key termsgradienty-intercept

Common mistake

Subtracting the coordinates in different orders on the top and bottom. Use $(y_2-y_1)$ and $(x_2-x_1)$ in the same order.

Section 2

The form y - y1 = m(x - x1)

If a line has gradient $m$ and passes through $(x_1,y_1)$ , its equation is $y-y_1=m(x-x_1).$ This is the quickest route when you are given a point and a gradient. Rearrange to $y=mx+c$ only if the question asks. Example: gradient $2$ through $(2,5)$ gives $y-5=2(x-2)$ , so $y=2x+1$ .

Key termspoint-gradient form

Common mistake

Writing $y-y_1=m(x+x_1)$ . If $x_1=-3$ the bracket is $(x+3)$ , not $(x-3)$ .

Exam tip

Check your final equation by substituting both given points.

Section 3

A line through two given points

To find the equation through $(x_1,y_1)$ and $(x_2,y_2)$ : first find $m=\frac{y_2-y_1}{x_2-x_1}$ , then use $y-y_1=m(x-x_1)$ with either point. Equivalently $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}.$ Example: $P(-3,4)$ , $Q(5,-2)$ gives $m=-\frac34$ and $y-4=-\frac34(x+3)$ .

Key termstwo-point form

Section 4

The form ax + by + c = 0

Many questions ask for the answer as $ax+by+c=0$ with integer $a$ , $b$ , $c$ . Multiply through to clear fractions and move every term to one side. From $y-4=-\frac34(x+3)$ : $4y-16=-3x-9$ , so $3x+4y-7=0$ . From this form the gradient is $-\frac{a}{b}$ and the $y$ -intercept is $-\frac{c}{b}$ (for $b\ne0$ ). Any integer multiple of the equation is also correct.

Key termsgeneral form

Common mistake

Reading the gradient of $3x+4y-7=0$ as $3$ or $\frac34$ . Rearrange first: $y=-\frac34x+\frac74$ .

Section 5

Intercepts and areas

A line meets the $x$ -axis where $y=0$ and the $y$ -axis where $x=0$ . For $3x+4y-7=0$ the intercepts are $\left(\frac73,0\right)$ and $\left(0,\frac74\right)$ . The triangle formed with the origin has area $\frac12\times\frac73\times\frac74=\frac{49}{24}$ . The midpoint of $(x_1,y_1)$ and $(x_2,y_2)$ is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ , which is often needed to build a line.

Key termsinterceptmidpoint

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Equation of a straight line

A line $l$ passes through the points $A(2,5)$ and $B(6,13)$ .
Find the coordinates of the point where $l$ meets the $x$ -axis.2 marks
The line $m$ has equation $3x-4y+8=0$ .
Find the $x$ -coordinate of the point on $m$ where $y=7$ .2 marks
A line $l_1$ passes through the points $P(-3,4)$ and $Q(5,-2)$ .
Find an equation of the line $l_1$ , giving your answer in the form $ax+by+c=0$ , where $a$ , $b$ and $c$ are integers.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).