OxfordAQA International AS and A-level Mathematics: Straight lines and intersections of lines with curves (9660)

This guide covers section P1.2 (Coordinate geometry) of the OxfordAQA International AS and A-level Mathematics specification (9660), Version 5.2, for International AS exams from May/June 2018 and A-level exams from May/June 2019. The specification places this content in International AS Unit P1 (Pure Maths); it is not split into core and extension parts, and it carries on into the A-level. It is also the algebra you rely on when you meet tangents in differentiation.

Pair it with the revision notes and the practice questions. For the rest of the unit see the Unit P1 overview, the course hub and the printable checklist. To find out where you stand first, try the free 10-minute diagnostics.

What P1.2 asks you to do

Specification content	What you must be able to do
Equation of a straight line	Write it as y − y₁ = m(x − x₁), as ax + by + c = 0 and as y = mx + c, and move between them
Problems with gradients, mid-points and distance	Find the gradient, mid-point and length of a line segment from two points
Parallel and perpendicular lines	Use equal gradients for parallel lines and a product of −1 for perpendicular lines
Intersection of a straight line and a curve	Solve algebraically, then interpret equal roots, distinct real roots and no real roots geometrically

Every method below is algebra you can do by hand, and the working is what shows your understanding.

Gradient, mid-point and distance

For points A(x₁, y₁) and B(x₂, y₂):

Quantity	Formula
Gradient of AB	m = (y₂ − y₁)/(x₂ − x₁)
Mid-point of AB	((x₁ + x₂)/2, (y₁ + y₂)/2)
Length AB	√[(x₂ − x₁)² + (y₂ − y₁)²]

The gradient is the change in y for a change of 1 in x. A positive gradient rises left to right, a negative one falls, and a vertical line has no gradient at all.

Worked example 1. A is (−3, 7) and B is (5, 1). Find the gradient, mid-point and length of AB.

Gradient: (1 − 7)/(5 − (−3)) = −6/8 = −3/4
Mid-point: ((−3 + 5)/2, (7 + 1)/2) = (1, 4)
Length: √(8² + (−6)²) = √(64 + 36) = √100 = 10

Keep brackets round negative coordinates. Writing 5 − −3 is how sign errors start.

Three forms of a straight line

Form	Best used when
y = mx + c	You know the gradient m and the y-intercept c, or want to read them off
y − y₁ = m(x − x₁)	You know the gradient and one point (x₁, y₁)
ax + by + c = 0	The answer must be in this form, usually with a, b, c whole numbers

The y − y₁ = m(x − x₁) form is the workhorse. Both the point and the gradient go in directly, with no need to find c first.

Worked example 2. Find the line through A(−3, 7) and B(5, 1), in the form ax + by + c = 0 with whole-number coefficients.

m = −3/4 (from example 1)
y − 7 = −3/4 (x + 3)
4(y − 7) = −3(x + 3)
4y − 28 = −3x − 9
3x + 4y − 19 = 0

Check by substituting the other point: 3(5) + 4(1) − 19 = 0. It works, so the line is right.

Worked example 3. Find the line through (2, −5) with gradient 3/2.

y + 5 = 3/2 (x − 2)
2y + 10 = 3x − 6
3x − 2y − 16 = 0

Multiply through by the denominator to clear fractions, then collect everything on one side. Keep the x term positive if you can.

Reading a gradient from ax + by + c = 0. Rearrange to y = −(a/b)x − c/b. The gradient is −a/b and the y-intercept is −c/b. For 6x − 4y + 9 = 0 the gradient is 6/4 = 3/2 and the intercept is 9/4. Beware the sign: the gradient is minus a over b, not a over b.

Parallel and perpendicular lines

• Parallel lines have equal gradients: m₁ = m₂.
• Perpendicular lines have gradients with product −1: m₁m₂ = −1. Equivalently, one gradient is the negative reciprocal of the other. Flip the fraction and change the sign.

Examples of negative reciprocals: 2/3 and −3/2, 5 and −1/5, −4/7 and 7/4.

Two warnings. First, compare gradients only after both lines are in a form where the gradient is visible. The lines 4x − 6y + 5 = 0 and 2x − 3y = 1 both have gradient 2/3, so they are parallel, even though the coefficients look different. Second, a horizontal line (gradient 0) is perpendicular to a vertical line (no gradient). The product rule cannot be used for that pair.

Worked example 4 (perpendicular bisector). Find the perpendicular bisector of AB, where A and B are as in example 1.

Gradient of AB = −3/4, so the perpendicular gradient is 4/3.
The bisector passes through the mid-point (1, 4).
y − 4 = 4/3 (x − 1)
3y − 12 = 4x − 4
4x − 3y + 8 = 0

Every point on this line is equidistant from A and B. A quick check: the point (1, 4) is the mid-point, and 4 − 12 + 8 = 0.

Worked example 5 (parallel line). Find the line through (6, −2) parallel to 5x − 2y = 7.

Rearrange: 2y = 5x − 7, so the gradient is 5/2.
y + 2 = 5/2 (x − 6)
2y + 4 = 5x − 30
5x − 2y − 34 = 0

A parallel line keeps the left-hand side, 5x − 2y, and changes only the constant. Here 5(6) − 2(−2) = 34, which gives the same answer in one line.

Where a line meets a curve

To find where a line meets a curve, solve their equations simultaneously. The method never changes.

1. Substitute the line (written as y = … or x = …) into the curve.
2. Rearrange into a quadratic, usually with everything on one side equal to zero.
3. Solve it by factorising, the formula or completing the square.
4. Put each x-value back into the line equation to get y. The line is easier to substitute into than the curve.
5. Write the answers as coordinate pairs.

Worked example 6. Find where y = 2x − 1 meets y = x² − 5x + 9.

x² − 5x + 9 = 2x − 1
x² − 7x + 10 = 0
(x − 2)(x − 5) = 0
x = 2 or x = 5
Points: (2, 3) and (5, 9)

For x = 2 the line gives y = 2(2) − 1 = 3, and for x = 5 it gives y = 9. Substituting into the curve returns the same values, and doing so is a worthwhile check.

Do not stop at the x-values. A question that asks for the points of intersection needs both coordinates.

What the discriminant tells you

After substitution you have a quadratic ax² + bx + c = 0. Its discriminant is b² − 4ac, and it decides how many times the line meets the curve.

Discriminant	Roots	Geometry
b² − 4ac > 0	Two distinct real roots	The line cuts the curve at two points
b² − 4ac = 0	Equal (repeated) roots	The line touches the curve at one point, so it is a tangent
b² − 4ac < 0	No real roots	The line and curve do not meet

Equal roots mean one point of contact, found from x = −b/(2a). This reading of the discriminant is exactly what P1.2 asks you to be able to interpret.

Worked example 7 (finding a tangent gradient). The line y = mx + 1 touches the curve y = x² + 2x + 5. Find the possible values of m and the points of contact.

x² + 2x + 5 = mx + 1
x² + (2 − m)x + 4 = 0

The line touches the curve when this has equal roots:

(2 − m)² − 4(1)(4) = 0
(2 − m)² = 16
2 − m = 4 or 2 − m = −4
m = −2 or m = 6

Points of contact, using x = −(2 − m)/2:

m = −2: x = −2, y = −2(−2) + 1 = 5   → (−2, 5)
m = 6:  x = 2,  y = 6(2) + 1 = 13    → (2, 13)

Both points lie on the curve: (−2)² + 2(−2) + 5 = 5 and 2² + 2(2) + 5 = 13. There are two tangents because there are two lines through (0, 1) that just touch the curve.

Worked example 8 (range of values). For the same family of lines, for what values of m does the line cut the curve at two points?

The discriminant is (2 − m)² − 16, which factorises as (m − 6)(m + 2). We need this greater than 0.

(m − 6)(m + 2) > 0
m < −2 or m > 6

The sketch of a positive U-shaped quadratic in m shows why: it is above zero outside its roots. For −2 < m < 6 the line misses the curve, and at m = −2 and m = 6 it is a tangent.

Worked example 9 (curve that is not a parabola). Curves like y = k/x also appear. Multiply through by x to get a quadratic. To find where y = 12/x meets 3x + y = 13:

3x + 12/x = 13
3x² + 12 = 13x
3x² − 13x + 12 = 0
(3x − 4)(x − 3) = 0
x = 4/3 or x = 3
y = 13 − 3x: y = 9 or y = 4
Points: (4/3, 9) and (3, 4)

The quadratic route works for any curve that becomes a quadratic after substitution. If the equation you get is a cubic, the same reading of repeated roots still applies, but you need to factorise it first.

Putting the ideas together

Questions often join several of these skills. A typical chain: find a gradient from two points, write the line, find a perpendicular bisector or a parallel line, then intersect it with a curve and interpret the discriminant. Work one step at a time, label each result, and keep exact values (fractions and surds) until the end.

Common errors

• Using y₂ − y₁ over x₁ − x₂, with the subtraction in different orders on top and bottom. The gradient of the line falling from (−3, 7) to (5, 1) is negative, so check the sign against a quick sketch.
• Forgetting the minus when reading the gradient from ax + by + c = 0.
• Taking the perpendicular gradient as just the reciprocal or just the negative. You need both flips.
• Giving only x-values after solving for the intersection.
• Setting the discriminant of the curve to zero rather than the discriminant of the combined quadratic. A tangent condition always comes from the equation after substitution.
• Writing b² − 4ac > 0 for a tangent. A tangent needs equals zero. Greater than zero means two points.
• Leaving the answer as y = (−3/4)x + 19/4 when the question asks for whole-number coefficients in the form ax + by + c = 0.
• Squaring a bracket wrongly: (2 − m)² is not 4 − m².

Official syllabus

This guide follows section P1.2 (Coordinate geometry) in section 3.1 of the OxfordAQA International AS and A-level Mathematics specification (9660), Version 5.2, published by OxfordAQA. Check the current specification on the OxfordAQA website for any updates.