E3.5 Equations of linear graphs

Learning objectives

E3.5.1Interpret and obtain the equation of a straight-line graph• Interpret and obtain the equation of a straight-line graph. Questions may: • use and request lines in different forms, e.g. ax + by = c y = mx + c x = k • involve finding the equation when the graph is given • ask for the gradient or y-intercept of a graph from an equation, e.g. find the gradient and y-intercept of the graph with equation 5x + 4y = 8. Candidates are expected to give equations of a line in a fully simplified form.

A straight-line equation links every point on the line. In $y=mx+c$ , $m$ is the gradient and $c$ is the $y$ -coordinate where the line crosses the $y$ -axis, so the intercept point is $(0,c)$ .

Form	How to interpret it
$y=mx+c$	read gradient $m$ and $y$ -intercept $(0,c)$ directly
$ax+by=c$ , $b\ne0$	rearrange to make $y$ the subject
$x=k$	vertical line through every point with $x$ -coordinate $k$ ; gradient undefined

$5x+4y=8\quad\Longrightarrow\quad y=-\frac54x+2\quad\Longrightarrow\quad m=-\frac54,\ (0,2)$

To obtain an equation from a graph or two points: find the gradient $m$ ; substitute one known point into $y=mx+c$ to calculate $c$ ; then write and fully simplify the equation. A graph may allow $c$ to be read directly from the $y$ -axis.

$A(3,16),\ B(8,31):\quad m=\frac{31-16}{8-3}=3,\quad 16=3(3)+c\Rightarrow c=7,\quad y=3x+7$

Substitute the other point: $3(8)+7=31$ , so both $A$ and $B$ satisfy the equation. When a requested form is specified, rearrange the final result into that form and remove common factors or unnecessary signs.

Do not confuse the intercept value $c$ with the point $(0,c)$ , and do not force a vertical line into $y=mx+c$ . Also distinguish $y=5-3x$ from $y=5x-3$ : coefficients and constants have different roles.

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—E3.6 Parallel linesRead