Gradient, intercepts, parallel/perpendicular conditions and the distance and midpoint formulas cover the content; the exam skill is choosing the right form (\(y - y_1 = m(x - x_1)\) or \(ax + by + c = 0\)) and using geometry (perpendicular bisectors, areas of triangles) rather than heavy algebra.
Equations and gradients (MM3.1, M4.10)
- Gradient \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\); line through \((x_1, y_1)\): \(y - y_1 = m(x - x_1)\).
- \(ax + by + c = 0\) has gradient \(-\frac ab\), \(x\)-intercept \(-\frac ca\) and \(y\)-intercept \(-\frac cb\); the intercept form is \(\frac xp + \frac yq = 1\).
- Parallel: equal gradients. Perpendicular: \(m_1 m_2 = -1\) (a line of gradient \(\frac ab\) has perpendicular gradient \(-\frac ba\)).
- Three points are collinear if two of the pairwise gradients agree.
- Midpoint \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\); distance \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
Useful geometry
- Perpendicular bisector of \(AB\): through the midpoint with the negative reciprocal gradient. Points equidistant from \(A\) and \(B\) lie on it.
- Distance from a point to a line \(ax + by + c = 0\): \(\dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}\) - or use 'area of triangle two ways'.
- Distance between parallel lines \(ax + by = c_1\) and \(ax + by = c_2\): \(\dfrac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}\).
- Area of a triangle with a horizontal or vertical base is \(\frac12 \times\) base \(\times\) height; a line through a vertex bisects the area iff it passes through the midpoint of the opposite side.
Must-know (no formula booklet)
- \(y - y_1 = m(x - x_1)\)
- perpendicular: \(m_1 m_2 = -1\)
- \(ax + by + c = 0\) has gradient \(-a/b\)
- distance from \((x_0,y_0)\) to \(ax+by+c=0\) is \(|ax_0+by_0+c|/\sqrt{a^2+b^2}\)
Classic traps
- A vertical line \(x = k\) has no gradient; do not divide by zero when points share an \(x\)-coordinate.
- When finding where \(y = mx + 4\) and \(y = 3x - 2\) meet 'in the first quadrant', both coordinates must be positive - two conditions on \(m\).
- Perpendicular to \(3x - 2y + 5 = 0\): gradient \(-\frac23\), not \(\frac23\) or \(-\frac32\).
Worked example
Q. Find the perpendicular bisector of \((1, 5)\) and \((7, 1)\).
A. Midpoint \((4, 3)\); segment gradient \(-\frac23\); bisector gradient \(\frac32\): \(y - 3 = \frac32(x - 4)\), i.e. \(3x - 2y = 6\).
How it has been examined
November 2017 · Paper 1Question 3area enclosed by two perpendicular lines and the x-axisquestion papersit online