ZIMSEC A Level · Pure Mathematics · Form 5
Straight lines
- Gradient m = (y₂ − y₁)/(x₂ − x₁); line: y − y₁ = m(x − x₁).
- Parallel lines: equal gradients. Perpendicular lines: m₁m₂ = −1.
- Midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2); distance √[(x₂ − x₁)² + (y₂ − y₁)²].
Circles
- Centre (a, b), radius r: (x − a)² + (y − b)² = r².
- General form x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g² + f² − c).
- A tangent is perpendicular to the radius at the point of contact.
Worked example: Find the centre and radius of x² + y² − 6x + 4y − 12 = 0.
(x − 3)² − 9 + (y + 2)² − 4 − 12 = 0 → (x − 3)² + (y + 2)² = 25
Centre (3, −2), radius 5.
Worked example: Find the tangent to that circle at the point (6, 2).
Gradient of radius from (3, −2) to (6, 2) = 4/3, so the tangent gradient = −3/4.
y − 2 = −¾(x − 6) → 4y − 8 = −3x + 18 → 3x + 4y = 26
Parametric equations
x = f(t), y = g(t). Eliminate t to get the Cartesian equation; dy/dx = (dy/dt) ÷ (dx/dt).
Exam tip: Draw a quick sketch – it helps you check signs and positions.
Check yourself
- Find the equation of the perpendicular bisector of A(1, 2) and B(5, 6).
- Find the equation of the circle with centre (−1, 4) through (2, 0).
- x = t², y = 2t. Find the Cartesian equation.
Show answers
- Midpoint (3, 4); gradient AB = 1, so perpendicular gradient −1: y = −x + 7
- r² = 9 + 16 = 25: (x + 1)² + (y − 4)² = 25
- t = y/2, so x = y²/4, i.e. y² = 4x
Questions about this topic (0)
No questions about this topic yet.
Sign in or create a free account to join in.