Straight‑line graphs
Learning objective
Use the equation of a straight line to solve problems involving gradients, intercepts, parallelism, perpendicularity, mid‑points, lengths, perpendicular bisectors and the linearisation of non‑linear data (Cambridge IGCSE Additional Mathematics 0606 – Section 7).
Key concepts
- The graph of a linear equation is a straight line.
- Gradient (slope) $m$ – the rate of change of $y$ with respect to $x$; $m=\dfrac{\Delta y}{\Delta x}$.
- $y$‑intercept $c$ – the value of $y$ when $x=0$; point $(0,c)$.
- $x$‑intercept $a$ – the value of $x$ when $y=0$; point $(a,0)$.
- A line is vertical when $x=k$ (gradient undefined) and horizontal when $y=c$ (gradient $0$).
- Two non‑vertical lines are parallel ⇔ their gradients are equal ($m_1=m_2$).
Both vertical lines are also parallel.
- Two non‑vertical lines are perpendicular ⇔ $m_1m_2=-1$.
A vertical line is perpendicular to any horizontal line.
Standard forms of a straight‑line equation
| Form |
Expression |
When it is useful |
| Slope‑intercept |
$y = mx + c$ |
Gradient $m$ and $y$‑intercept $c$ are known. |
| Point‑slope |
$y - y_1 = m\,(x - x_1)$ |
Gradient $m$ and a single point $(x_1,y_1)$ are known. |
| Two‑point form |
$\displaystyle\frac{y-y_1}{x-x_1}= \frac{y_2-y_1}{x_2-x_1}$ |
Two points $(x_1,y_1)$ and $(x_2,y_2)$ are given. |
| Intercept form |
$\displaystyle\frac{x}{a} + \frac{y}{c}=1$ |
$x$‑intercept $a$ and $y$‑intercept $c$ are known. |
Finding the gradient
If two points $(x_1,y_1)$ and $(x_2,y_2)$ lie on the line,
\[
m = \frac{y_2-y_1}{\,x_2-x_1\,}
\]
Finding intercepts from an equation
- Set $x=0$ → the resulting $y$ is the $y$‑intercept $c$.
- Set $y=0$ → solve for $x$ to obtain the $x$‑intercept $a$.
Parallel and perpendicular lines – special cases
- Parallel: $m_1=m_2$ (or both lines are vertical, $x=k_1$ and $x=k_2$).
- Perpendicular: $m_1m_2=-1$ (provided neither line is vertical).
A vertical line $x=k$ is perpendicular to any horizontal line $y=c$.
Worked example – parallel & perpendicular
Line $L_1$: $y = 3x - 4$.
Line $L_2$ passes through $(2,1)$ and is perpendicular to $L_1$.
- Gradient of $L_1$ is $m_1 = 3$.
- For $L_2$, $m_2 = -\dfrac{1}{m_1}= -\dfrac13$.
- Point‑slope with $(2,1)$: $y-1 = -\dfrac13\,(x-2)$.
- Convert to slope‑intercept: $y = -\dfrac13x + \dfrac73$.
Worked example – using intercepts
Find the equation in intercept form of the line that cuts the axes at $(4,0)$ and $(0,-3)$.
- Here $a=4$ and $c=-3$.
- Intercept form: $\displaystyle\frac{x}{4} + \frac{y}{-3}=1$.
- Multiply by $12$ to obtain a more familiar form: $3x - 4y = 12$.
- In slope‑intercept form: $y = \frac{3}{4}x - 3$ (gradient $m=\tfrac34$, $y$‑intercept $-3$).
Mid‑point, length and perpendicular bisector
- Mid‑point of $AB$ (with $A(x_1,y_1)$, $B(x_2,y_2)$):
\[
M\Bigl(\frac{x_1+x_2}{2},\;\frac{y_1+y_2}{2}\Bigr)
\]
- Length of $AB$** (distance formula):
\[
|AB| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
\]
- Perpendicular bisector of $AB$**:
- Find the gradient $m_{AB}= \dfrac{y_2-y_1}{x_2-x_1}$.
- The bisector’s gradient $m_{\perp}= -\dfrac{1}{m_{AB}}$ (if $AB$ is vertical, the bisector is horizontal, $m_{\perp}=0$).
- Use the mid‑point $M$ and point‑slope form:
\[
y - y_M = m_{\perp}\,(x - x_M)
\]
Worked example – perpendicular bisector
Find the equation of the perpendicular bisector of the segment joining $A(1,2)$ and $B(5,8)$.
- Mid‑point $M\bigl(\tfrac{1+5}{2},\tfrac{2+8}{2}\bigr) = (3,5)$.
- Gradient of $AB$: $m_{AB}= \dfrac{8-2}{5-1}= \dfrac{6}{4}= \dfrac32$.
- Gradient of the bisector: $m_{\perp}= -\dfrac{1}{\frac32}= -\frac23$.
- Point‑slope through $M(3,5)$: $y-5 = -\frac23\,(x-3)$.
- Slope‑intercept form: $y = -\frac23x + \frac{19}{3}$.
Linearisation (transforming non‑linear relationships to straight lines)
Many practical relationships are of the form $y = A x^{n}$, $y = A b^{x}$ or $y = A\ln x + B$. By applying logarithms or roots the data can be plotted as a straight line, allowing the use of the methods above to determine the unknown constants.
| Original relation |
Transformation |
Resulting straight‑line form |
What the graph gives |
| $y = A x^{n}$ |
Take $\log$ of both sides (any base, usually 10 or $e$) |
$\log y = n\log x + \log A$ |
Slope $=n$, intercept $=\log A$ (log–log plot). |
| $y = A b^{x}$ |
Take $\log$ of both sides |
$\log y = x\log b + \log A$ |
Slope $=\log b$, intercept $=\log A$ (semi‑log plot). |
| $y = A\ln x + B$ |
Plot $y$ against $\ln x$ (no algebraic rearrangement needed) |
$y = A(\ln x) + B$ |
Slope $=A$, intercept $=B$ (semi‑log plot with natural log). |
Worked example – linearising a power law
Experimental data suggest $y = k x^{3}$. The points recorded are:
Transform to a log–log plot (base 10 shown):
| $x$ | $y$ | $\log x$ | $\log y$ |
|---|
| 1 | 2 | 0.0000 | 0.3010 |
| 2 | 16 | 0.3010 | 1.2041 |
| 3 | 54 | 0.4771 | 1.7324 |
Using any two points, the gradient $n$ is
\[
n = \frac{1.2041-0.3010}{0.3010-0.00}= \frac{0.9031}{0.3010}=3.00
\]
Thus the exponent is $3$. From $\log y = n\log x + \log A$, using the point $(0,0.3010)$ gives $\log A =0.3010$, so $A = 2$. The final equation is $y = 2x^{3}$.
Full synthesis example
Find the gradient, $y$‑intercept, $x$‑intercept and equation (slope‑intercept form) of the line through $A(2,5)$ and $B(6,13)$. State whether the line $3y-9x=12$ is parallel, perpendicular or neither to this line.
- Gradient:
\[
m = \frac{13-5}{6-2}= \frac{8}{4}=2
\]
- Point‑slope using $A$:
\[
y-5 = 2(x-2)
\]
- Slope‑intercept:
\[
y = 2x + 1
\]
- $y$‑intercept $c = 1$ (point $(0,1)$).
- $x$‑intercept: set $y=0$ → $0=2x+1$ → $x=-\tfrac12$ (point $(-\tfrac12,0)$).
- For $3y-9x=12$, rewrite:
\[
y = 3x + 4\quad\Rightarrow\quad m' = 3
\]
Compare gradients: $m=2$, $m'=3$. They are not equal and $2\times3=6eq-1$, so the lines are neither parallel nor perpendicular.
Typical exam questions
- Given $3y-9x=12$, find the gradient, $y$‑intercept and $x$‑intercept. Write the equation in slope‑intercept form.
- The line $L$ has gradient $-4$ and passes through $(3,2)$.
- Write the equation of $L$ in slope‑intercept form.
- State the coordinates of its $y$‑intercept and $x$‑intercept.
- Two lines are $y=\tfrac12x+3$ and $y=-2x+k$.
- Find $k$ if the lines intersect at $(4,5)$.
- State the gradient of each line and comment on whether they are parallel, perpendicular or neither.
- A line passes through $P(-2,4)$ and $Q(4,-2)$.
- Find the equation of the line in intercept form.
- State the $x$‑intercept and $y$‑intercept.
- Find the equation of the perpendicular bisector of the segment joining $A(1,2)$ and $B(5,8)$.
- Experimental data are thought to follow $y = A b^{x}$. The table below is given. Linearise the data, determine $b$ and $A$, and write the final equation.
Summary checklist
- Identify which form of the linear equation you are given.
- Calculate the gradient:
\[
m = \frac{y_2-y_1}{x_2-x_1}
\]
(or read it directly from $y=mx+c$).
- Find intercepts by setting $x=0$ (for $c$) and $y=0$ (for $a$).
- For two lines:
- Parallel ⇔ $m_1=m_2$ (or both vertical).
- Perpendicular ⇔ $m_1m_2=-1$ (or one vertical and the other horizontal).
- Mid‑point: $\bigl(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigr)$.
Length: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
- Perpendicular bisector: use the negative reciprocal of the segment’s gradient through the mid‑point.
- Linearise non‑linear relationships using the appropriate log or root transformation, then apply the straight‑line methods to find unknown constants.
- Always check your final equation by substituting known points or constants.