amazon.co.ukStraight lines, gradients, parallel and perpendicular conditions, circle equations, tangents, chord lengths and loci - A-Level pure maths coordinate geometry (Edexcel 9MA0 Topic 3, AQA 7357 Section C) revised with worked examples and mark-scheme logic.
Straight-line work underpins half the pure paper (Edexcel 9MA0 3.1; AQA 7357 C1; OCR A H640 4.01; WJEC/Eduqas and CCEA AS Pure Mathematics). The gradient between two points is m = (y2 - y1)/(x2 - x1), and the equation of a line comes in two interchangeable forms: point-slope y - y1 = m(x - x1) and slope-intercept y = mx + c. Exam answers usually prefer ax + by + c = 0 form when a question says give your answer in the form ax + by + c = 0 - multiply through and collect every term on one side.
Two conditions do the geometric work: parallel lines have equal gradients (m1 = m2) and perpendicular lines have gradients whose product is -1 (m1 x m2 = -1), so a perpendicular gradient is the negative reciprocal. For a line 2x + 3y = 6, rewrite as y = -2x/3 + 2 to read the gradient -2/3; any perpendicular line has gradient 3/2.
Find the equation of the perpendicular bisector of the line joining A(-1, 4) and B(5, -2).
Line 1: Midpoint of AB = ((-1 + 5)/2, (4 - 2)/2) = (2, 1) (M1).
Line 2: Gradient of AB = (-2 - 4)/(5 - (-1)) = -6/6 = -1 (M1).
Line 3: Perpendicular gradient = -1/(-1) = 1 (A1).
Line 4: Through (2, 1): y - 1 = 1(x - 2), so y = x - 1 (A1).
Check the answer: any point on y = x - 1 should be equidistant from A and B - testing (2, 1) gives distance sqrt(9 + 9) = sqrt(18) to both, confirming the bisector.
The circle with centre (a, b) and radius r has equation (x - a)^2 + (y - b)^2 = r^2 (Edexcel 9MA0 3.6; AQA 7357 C2; OCR A H640 5.01; CCEA AS1/A21). Two directions of travel: (1) given centre and radius, write the equation; (2) given the expanded form x^2 + y^2 + 2gx + 2fy + c = 0, complete the square in x and in y to recover the centre (-g, -f) and radius sqrt(g^2 + f^2 - c). Centre and radius questions award method marks for the completing-the-square lines, so show them.
A circle can be specified without its centre: as the circle on a given diameter, where the centre is the midpoint of the diameter's endpoints and the radius is half the distance between them. This construction appears constantly in exam questions.
Find the equation of the circle with diameter endpoints A(1, 2) and B(5, 6), and find the tangent at A.
Line 1: Centre = midpoint of AB = ((1 + 5)/2, (2 + 6)/2) = (3, 4) (M1).
Line 2: Radius = half of AB: AB = sqrt((5 - 1)^2 + (6 - 2)^2) = sqrt(32), so r^2 = 32/4 = 8 (M1 A1).
Line 3: Circle: (x - 3)^2 + (y - 4)^2 = 8 (A1).
Line 4: Gradient of the radius from centre (3, 4) to A(1, 2): (2 - 4)/(1 - 3) = 1 (M1).
Line 5: Tangent gradient = -1 (perpendicular condition). Tangent at A: y - 2 = -1(x - 1), so y = -x + 3 (A1).
Note the efficient habit: keep r^2 as a value rather than computing r, since the equation only needs r^2. The angle in a semicircle is a right angle (Thales), which is why a diameter-endpoint question often asks to prove a triangle is right-angled.
Three circle properties are examinable (Edexcel 9MA0 3.6; AQA 7357 C2; OCR A H640 5.01): (1) the tangent is perpendicular to the radius at the point of contact - the workhorse for tangent equations; (2) the perpendicular bisector of a chord passes through the centre - the workhorse for finding a centre from a chord; (3) the angle in a semicircle is 90 degrees. Tangency can also be established algebraically: substitute the line into the circle, and a repeated root (discriminant zero) proves the line touches at exactly one point.
Show that the line 3x + 4y = 25 is a tangent to the circle x^2 + y^2 = 25, and find the point of contact.
Line 1: Substitute x = (25 - 4y)/3: ((25 - 4y)/3)^2 + y^2 = 25 (M1: substitution).
Line 2: Multiply by 9: (25 - 4y)^2 + 9y^2 = 225, so 625 - 200y + 16y^2 + 9y^2 = 225 (M1).
Line 3: 25y^2 - 200y + 400 = 0, so y^2 - 8y + 16 = 0, i.e. (y - 4)^2 = 0 (A1: repeated root).
Line 4: The discriminant is zero, so there is exactly one intersection point: the line is a tangent (A1: conclusion).
Line 5: Point of contact: y = 4, x = (25 - 16)/3 = 3, i.e. (3, 4) (A1).
Find the tangent to the circle (x - 2)^2 + (y + 1)^2 = 25 at the point (5, 3).
Line 1: Radius from centre (2, -1) to (5, 3): gradient (3 - (-1))/(5 - 2) = 4/3 (M1).
Line 2: Tangent gradient = -3/4 (A1: perpendicular).
Line 3: y - 3 = -(3/4)(x - 5), so 4y - 12 = -3x + 15, i.e. 3x + 4y = 27 (A1).
To find a centre from a chord: the centre lies on the chord's perpendicular bisector - combine with a second chord (or a point and the radius) to pin down the centre by intersecting two bisectors.
Intersections are simultaneous equation problems. Line with line: solve the pair directly. Line with circle: substitute the line's expression for y (or x) into the circle to get a quadratic in one variable; its roots give the intersection points. Two outcomes matter: two distinct roots mean the line cuts the circle (a chord), a repeated root means tangency, and no real roots mean the line misses.
Chord lengths can be found without ever computing both points in full: if the quadratic in x has roots x1 and x2 and the line has gradient m, then the chord length is sqrt(1 + m^2) |x1 - x2| - but in practice, finding both points and using the distance formula is safer under exam pressure.
The line y = 2x + 1 intersects the circle (x - 1)^2 + (y - 2)^2 = 10 at points A and B. Find the exact length AB.
Line 1: Substitute: (x - 1)^2 + (2x + 1 - 2)^2 = 10, so (x - 1)^2 + (2x - 1)^2 = 10 (M1).
Line 2: x^2 - 2x + 1 + 4x^2 - 4x + 1 = 10, giving 5x^2 - 6x - 8 = 0 (A1).
Line 3: Factorising: (5x + 4)(x - 2) = 0, so x = 2 or x = -4/5 (A1).
Line 4: Points: A(2, 5) and B(-4/5, -3/5) (M1: y = 2x + 1 for each).
Line 5: AB = sqrt((2 + 4/5)^2 + (5 + 3/5)^2) = sqrt((14/5)^2 + (28/5)^2) = sqrt(980/25) = sqrt(980)/5 = 14sqrt(5)/5 (A1: exact form).
Keep every step exact: fractions, surds and squared terms should stay unevaluated until the final line, since rounding early destroys exact answers and drops accuracy marks. When the quadratic in x does not factorise nicely, use the quadratic formula for the two roots and continue exactly as before - the chord length method does not depend on the roots being rational.
A locus question translates a geometric condition into an equation. Equidistance from two points gives the perpendicular bisector; equidistance from a point and a line gives a parabola. The method: write the distance to each object, equate, square to remove the roots, and simplify.
Find the locus of points equidistant from the point (2, 3) and the line y = -1.
Line 1: Distance to the point: sqrt((x - 2)^2 + (y - 3)^2) (M1).
Line 2: Distance to the line y = -1: |y + 1| (M1).
Line 3: Equate and square: (x - 2)^2 + (y - 3)^2 = (y + 1)^2 (A1).
Line 4: Expand: (x - 2)^2 + y^2 - 6y + 9 = y^2 + 2y + 1, so (x - 2)^2 = 8y - 8 (A1).
Line 5: y = 1 + (x - 2)^2/8: a parabola with vertex (2, 1), focus (2, 3) and directrix y = -1 (A1).
Geometric proof problems use gradients and distances: to prove a triangle is right-angled, either show two gradients multiply to -1 or show the sides satisfy Pythagoras (a^2 + b^2 = c^2 with the longest side as c). For A(1, 1), B(2, 2) and C(4, 0): gradient of AB is 1, gradient of BC is (0 - 2)/(4 - 2) = -1, and 1 x (-1) = -1, so the angle at B is 90 degrees.
Exam technique: (1) write the completing-the-square or gradient lines explicitly - method marks live there; (2) keep r^2 unradiquified; (3) when a question says show that a line is a tangent, the expected proof is a zero discriminant, not a sketch; (4) use exact fractions throughout; (5) finish circle questions by checking the point lies on the circle (substitute into the equation). Board coverage: Edexcel 9MA0 Topic 3, AQA 7357 C1-C2, OCR A H640 4.01 and 5.01, WJEC/Eduqas AS Unit 1 geometry, CCEA AS1/A21 - identical techniques, differing wording.