Learn the method, study an example, then explain your reasoning in an exam-style problem. Work on paper before opening the solution and compare each step with the marking guidance.
Parallel, perpendicular and exceptional lines
Compare gradients while handling horizontal and vertical lines correctly.
Notes
Before you start: rearrange a line equation and work with fractions.
For two distinct points with different x-coordinates, the gradient is \(m=(y_2-y_1)/(x_2-x_1)\). Distinct nonvertical lines are parallel when their gradients agree. Equal gradients and equal intercepts give the same line, rather than two distinct parallel lines.
For two nonvertical perpendicular lines, \(m_1m_2=-1\). Thus the perpendicular gradient is \(-1/m\) when \(m\ne0\). Both the sign and reciprocal matter.
A vertical line has equation \(x=c\), with undefined gradient. Its perpendicular lines are horizontal, \(y=d\), with gradient zero. Do not apply the reciprocal formula to zero or call a vertical gradient infinity.
Recall later
A line perpendicular to gradient \(-3/4\) has gradient \(4/3\). A line perpendicular to \(x=7\) is horizontal.
Common mistakes
Taking only the reciprocal and forgetting the negative sign.
Using m₁m₂=−1 for a vertical/horizontal pair.
Calling coincident lines distinct parallel lines.
Three worked examples
Example 1
Find the gradient through \((-1,2),(3,8)\), and compare it with \(y=\frac32x-5\).
Show the reasoning
\(m=(8-2)/(3-(-1))=6/4=3/2\). The first line is \(y=\frac32x+\frac72\); the intercepts differ, so the two lines are distinct and parallel.
Example 2
Find the line through \((4,3)\) perpendicular to \(y=2x+1\).
Show the reasoning
The required gradient is \(-1/2\). Hence \(y-3=-\frac12(x-4)\), or \(y=-\frac12x+5\).
Example 3
Find the line through \((1,5)\) perpendicular to \(x=-2\).
Show the reasoning
The given line is vertical. The required line is horizontal through y=5, so its equation is \(y=5\).
