2.1 Equation of a straight lineIB Maths: Applications and Interpretation SL: Revision notes

The gradient $m$ of a line measures how steep it is: the change in $y$ for each one-unit change in $x$. For two points $(x_1,y_1)$ and $(x_2,y_2)$: $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\text{rise}}{\text{run}}.$ A positive gradient slopes upwards (left to right), a negative gradient slopes downwards, a horizontal line has $m=0$, and a vertical line has an undefined gradient. Example: $A(1,4)$ and $B(5,12)$ give $m=\frac{12-4}{5-1}=2$: $y$ rises by 2 for every 1 across.

There are three forms you must recognise and be able to convert between.

• Gradient-intercept form: $y=mx+c$, where $m$ is the gradient and $c$ the $y$-intercept.
• Point-gradient form: $y-y_1=m(x-x_1)$, for a line with gradient $m$ through $(x_1,y_1)$.
• General form: $ax+by+d=0$, usually with integer $a$, $b$, $d$. Example: through $A(1,4)$ with $m=2$: $y-4=2(x-1)$, so $y=2x+2$, or in general form $2x-y+2=0$. To find the gradient from general form, rearrange: $3x+4y-24=0\Rightarrow y=-\frac34x+6$, so $m=-\frac34$. In general $m=-\frac ab$.

The $y$-intercept is where the line crosses the $y$-axis, so set $x=0$. The $x$-intercept is where it crosses the $x$-axis, so set $y=0$. For $3x+4y-24=0$: $x=0$ gives $y=6$, and $y=0$ gives $x=8$. The intercepts are $(0,6)$ and $(8,0)$. In $y=mx+c$ the $y$-intercept is read straight from $c$. The $x$-intercept (the zero of the function) is $-\frac cm$.

Two lines with gradients $m_1$ and $m_2$ are:

• parallel if $m_1=m_2$ (they never meet);
• perpendicular if $m_1\times m_2=-1$, i.e. $m_2=-\frac{1}{m_1}$ (flip the fraction and change the sign). Example: a line perpendicular to $y=-\frac25x+1$ has gradient $\frac52$. Worked example: find the line through $(2,5)$ perpendicular to $y=2x+1$. Gradient $-\frac12$, so $y-5=-\frac12(x-2)$, giving $y=-\frac12x+6$, or $x+2y-12=0$.

Gradients of real slopes use the same rule: gradient $=\frac{\text{vertical rise}}{\text{horizontal distance}}$, with both lengths in the same units. A mountain road that rises 150 m over a horizontal distance of 2500 m has gradient $\frac{150}{2500}=0.06$, which is a 6% incline: 6 m of climb per 100 m horizontally. The same idea applies to a bridge ramp or a wheelchair access slope. The angle of an incline is $\theta=\tan^{-1}(m)$. In a linear model $y=mx+c$, the gradient is the rate of change and has units 'units of $y$ per unit of $x$' (e.g. 60 m per km); $c$ is the starting value when $x=0$.