Equations of Straight Lines

Slope

The slope of the line through $(x_1, y_1)$ and $(x_2, y_2)$ is

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

• Parallel lines: $m_1 = m_2$.
• Perpendicular lines: $m_1 m_2 = -1$.

Forms of the equation

Form	Equation
point-slope	$y - y_1 = m(x - x_1)$
two-point	$\dfrac{y - y_1}{x - x_1} = \dfrac{y_2 - y_1}{x_2 - x_1}$
slope-intercept	$y = mx + c$
general form	$Ax + By + C = 0$

From the general form $Ax + By + C = 0$:

$$\text{slope} = -\frac{A}{B}, \quad x\text{-intercept} = -\frac{C}{A}, \quad y\text{-intercept} = -\frac{C}{B}$$

Common mistake: the slope of $3x - 4y + 12 = 0$ is $-\frac{3}{-4} = \frac{3}{4}$. Watch the two minus signs.

Distance, mid-point and point of division

For $A(x_1, y_1)$ and $B(x_2, y_2)$:

• Distance $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
• Mid-point $= \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$
• $P$ divides $AB$ internally in the ratio $r : s$:

$$P = \left(\frac{s x_1 + r x_2}{r + s}, \frac{s y_1 + r y_2}{r + s}\right)$$

Note the cross-over: $s$ goes with $A$ and $r$ goes with $B$.

Intersection of two lines

Solve the two equations simultaneously.

Locus

A locus is the path traced by a moving point $P$ that satisfies a condition:

Condition on $P$	Locus
A fixed distance $r$ from a point $C$	A circle with centre $C$ and radius $r$
Equidistant from two points $A$ and $B$	The perpendicular bisector of $AB$
Equidistant from two parallel lines	A line parallel to both, midway between them
Equidistant from two intersecting lines	The pair of angle bisectors

To find the equation, let $P = (x, y)$, write the condition with the distance formula, then simplify.

Perpendicular bisector

The perpendicular bisector of $AB$ passes through the mid-point of $AB$ and has slope $-\dfrac{1}{m_{AB}}$. Every point on it is equidistant from $A$ and $B$.

Exam tip: give the final answer in general form $Ax + By + C = 0$ with integer coefficients unless the question asks otherwise.