Study Mathematics notes on Straight Line Graphs for A-Level. Useful for revision, homework, and exam preparation in United Kingdom.
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Straight Line Graphs Cheat Sheet
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Edexcel Pure Year 1
$\boldsymbol{y}=\boldsymbol{m} \boldsymbol{x}+\boldsymbol{c}$
The gradient $m$ of straight-line graphs can be found by taking two random points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ on the line and considering the horizontal and vertical distance between these two points.
The formula to calculate any given straight line's gradient is:
- $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Straight line equations come in the form of
- $y=m x+c$, where $m$ is the gradient and $c$ is the $y$ intercept.
Or
- $a x+b y+c=0$, where $a, b$ \& $c$ are all real numbers.
Example 1: Calculate the gradient of the following line. Point A has coordinate $(4,3)$ and point B has coordinate $(7,5)$. The line passes through both point A and B.
To calculate any gradient, we need know any two points which the line passes through. We are given two points, point $A(4,3)$ and point $B(7,5)$. The vertical distance between point $A$ and $B$ needs to be divided by the horizontal distance between the points.
- $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-3}{7-4}=\frac{2}{3}$
Example 2: A line $l$ has gradient 2 and $y$ intercept at $(0,-7)$. The line has equation $a x+b y+c=0$. Find the values of $a, b$ and $c$.
Straight line equations come in the form of $y=m x+c$, where $m$ is the gradient and $c$ is the $y$ intercept. In the question we are given that the gradient is 2 , hence $m=2$, and that the $y$ intercept is at ( $0,-7$ ), hence $c=-7$. Substituting these values into $y=m x+c$ we get
\[
y=2 x-7
\]
However, the question asks us to find the values of $a, b$ and $c$ on the form of $a x+b y+c=0$. Hence, we need to rearrange our equation into this form.
\[
2 x-y-7=0
\]
Therefore, the solutions to this question are $a=2, b=-1$ and $c=-7$.
Equations of Straight Lines
Instead of relying on using $y=m x+c$ to find the equation of a line, in some cases it may be useful to find the equation of a line using the following method.
If you know one point on the line and the gradient or two distinct points on the line, we can find the equation of the line by using the formula:
- $y-y_{1}=m\left(x-x_{1}\right)$
Example 3: Find the equation of the line that passes through the points $(7,2)$ and $(8,5)$.
First, we need to work out the gradient $m$
\[
m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-2}{8-7}=\frac{3}{1}=3
\]
We can take the point $(7,2)$ or $(8,5)$ to use in the …