Coordinate geometry, vectors and binomial expansion in depth

4.1 · Free notes

Straight lines: equations, gradients, parallel and perpendicular lines

Key points

• The gradient of the line through (x₁, y₁) and (x₂, y₂) is m = (y₂ − y₁)/(x₂ − x₁), which is change in y divided by change in x. Always subtract in the same order on top and bottom.
• The gradient-intercept form is y = mx + c, where m is the gradient and c is the y-intercept. Rearrange to this form to read off the gradient.
• The point-gradient form is y − y₁ = m(x − x₁). It is the quickest way to write down a line when you know one point and the gradient.
• The general form is ax + by + c = 0, where a, b and c are usually required to be integers. If a question asks for this form, clear any fractions before you give your answer.
• For a line written as ax + by + c = 0, the gradient is −a/b. Check that you have not forgotten the minus sign.
• Parallel lines have equal gradients (m₁ = m₂). A line parallel to a given line therefore keeps the same gradient and changes only the point it passes through.
• Perpendicular lines satisfy m₁ × m₂ = −1, so one gradient is the negative reciprocal of the other (m → −1/m). The gradient 2/3 becomes −3/2.
• A horizontal line y = k has gradient 0. A vertical line x = k has an undefined gradient. These two are perpendicular to each other, and the m₁m₂ = −1 rule does not apply to them.
• The gradient equals tan θ, where θ is the angle the line makes with the positive x-axis. For example, m = −1 gives θ = 135°, measured anticlockwise.
• Three points are collinear if the gradients between each pair are equal. To find an unknown coordinate, set the two gradients equal and solve.
• In context problems, the gradient is the rate of change (for example £ per km) and the intercept is the fixed or starting value. State both with their units.
• Always check by substitution: a point lies on a line only if its coordinates satisfy the equation. Substituting a given point back in is a quick way to catch sign errors.

In more detail

• Worked two points: (−1, 4) and (3, −4) → m = −8 ÷ 4 = −2 → y − 4 = −2(x + 1) → y = −2x + 2, or 2x + y − 2 = 0.
• Worked perpendicular: perpendicular to y = −2x + 2 through (4, 1) → m = ½ → y − 1 = ½(x − 4) → x − 2y − 2 = 0.
• Worked checking perpendicular: 3x − 2y = 5 (m = 3/2) and 2x + 3y = 1 (m = −2/3) → 3/2 × (−2/3) = −1, so they are perpendicular.
• Worked model: C = 25 + 0.12m → a fixed charge of £25 plus 12p per mile.