2.1 Equations of straight linesIB Maths: Analysis and Approaches SL: Revision notes

• Gradient–intercept form: $y = mx + c$, where $m$ is the gradient and $c$ the $y$-intercept.
• General form: $ax + by + d = 0$. Examiners often ask for integer $a$, $b$, $d$.
• Point–gradient form: $y - y_1 = m(x - x_1)$, for a line of gradient $m$ through $(x_1, y_1)$.

All three describe the same line; rearrange between them as needed. For $3x - 4y + 12 = 0$: $4y = 3x + 12$, so $y = \frac34x + 3$.

The gradient through $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$ — change in $y$ over change in $x$. A positive gradient slopes up to the right; a negative one slopes down.

The $y$-intercept is found by setting $x = 0$; the $x$-intercept by setting $y = 0$. For $y = -2x + 1$: $y$-intercept $1$, $x$-intercept $\frac12$.

Lines with gradients $m_1$ and $m_2$ are

• parallel if $m_1 = m_2$;
• perpendicular if $m_1\times m_2 = -1$, i.e. $m_2 = -\frac{1}{m_1}$ (the negative reciprocal).

The perpendicular to $y = \frac34x + 3$ has gradient $-\frac43$. Through $(6, -1)$: $y + 1 = -\frac43(x - 6)$, or $4x + 3y - 21 = 0$.

• A point lies on a line if its coordinates satisfy the equation.
• Two lines meet where their equations hold simultaneously: solve them together.
• The shortest distance from a point to a line is measured along the perpendicular through the point. Find the foot of the perpendicular (the intersection) and use the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Real inclines are described by their gradient as a decimal, a ratio or a percentage. A road rising 84 m over a horizontal 1200 m has gradient $\frac{84}{1200} = 0.07$, i.e. $7\%$. Always use the horizontal distance and consistent units. In a model $y = mx + c$, $m$ is the rate of change (metres of height per metre travelled horizontally) and $c$ is the starting value.