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

These are condensed notes for 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 content sits in International AS Unit P1 (Pure Maths) and is not split into core and extension parts. For full explanations and longer worked examples, read the study guide.

Then test yourself with the practice questions. Other useful pages are the Unit P1 revision notes, the course hub, the printable checklist and the free 10-minute diagnostics.

Formulas to know

Item	Result
Gradient through (x₁, y₁), (x₂, y₂)	m = (y₂ − y₁)/(x₂ − x₁)
Mid-point	((x₁ + x₂)/2, (y₁ + y₂)/2)
Distance	√[(x₂ − x₁)² + (y₂ − y₁)²]
Line, gradient and point	y − y₁ = m(x − x₁)
Line, gradient and intercept	y = mx + c
General form	ax + by + c = 0, gradient −a/b, y-intercept −c/b
Parallel	m₁ = m₂
Perpendicular	m₁m₂ = −1
Discriminant of ax² + bx + c = 0	b² − 4ac

Method boxes

Find the equation of a line

1. Get the gradient (given, or from two points).
2. Pick one known point.
3. Write y − y₁ = m(x − x₁).
4. If asked for ax + by + c = 0, clear fractions and collect terms on one side. Check with the second point.

Perpendicular bisector of AB

1. Mid-point M of AB.
2. Gradient of AB, then flip and change the sign.
3. Line through M with that gradient.

Intersection of a line and a curve

1. Substitute the line into the curve.
2. Rearrange to a quadratic equal to zero.
3. Solve.
4. Find y from the line.
5. Give coordinate pairs.

Tangent condition

1. Substitute and rearrange so the right side is 0.
2. Find b² − 4ac in terms of the unknown.
3. Set it equal to 0 for a tangent.
4. Solve, then find the point of contact from x = −b/(2a).

Discriminant and geometry

b² − 4ac	Roots	Line and curve
> 0	Two distinct real	Two points of intersection
= 0	Equal	One point, a tangent
< 0	None	No intersection

Small worked reminders

Perpendicular bisector. A(2, −3), B(10, 3).

Gradient AB = 6/8 = 3/4
Mid-point = (6, 0)
Perpendicular gradient = −4/3
y − 0 = −4/3 (x − 6)
3y = −4x + 24
4x + 3y − 24 = 0

Is the line a tangent? Does y = 2x − 1 touch y = x² − 2x + 3?

x² − 2x + 3 = 2x − 1
x² − 4x + 4 = 0
(x − 2)² = 0

The roots are equal, so yes. The point of contact has x = 2, y = 3, so it is (2, 3).

Family of lines. For the line y = x + k and the curve y = x² − 3x + 6:

x² − 4x + (6 − k) = 0
Discriminant: 16 − 4(6 − k) = 4k − 8

k	Discriminant	Result
k > 2	positive	two intersections
k = 2	zero	tangent at x = 2
k < 2	negative	none

A reciprocal curve. Where does y = 20/x meet 2x + y = 13?

2x + 20/x = 13
2x² − 13x + 20 = 0
(2x − 5)(x − 4) = 0
x = 5/2 or x = 4
y = 13 − 2x: y = 8 or y = 5
Points: (5/2, 8) and (4, 5)

A repeated factor in any polynomial, not just a quadratic, means the line touches the curve there.

Right angle and collinear checks. Use gradients.

A(−2, 1), B(4, 3), C(2, 9)
Gradient AB = 2/6 = 1/3
Gradient BC = 6/(−2) = −3
Product = −1, so AB is perpendicular to BC (right angle at B).

P(−1, −5), Q(2, 1), R(5, 7)
Gradient PQ = 6/3 = 2, gradient QR = 6/3 = 2
Same gradient and a shared point Q, so P, Q, R lie on one line.

Tangent with an unknown constant. The line y = 5x + c touches y = x² − x + 6. Find c and the point of contact.

x² − x + 6 = 5x + c
x² − 6x + (6 − c) = 0
Discriminant: 36 − 4(6 − c) = 12 + 4c = 0, so c = −3
Equal roots at x = −(−6)/2 = 3
y = 5(3) − 3 = 12, so the contact point is (3, 12)

Range of gradients. For which m does y = mx − 3 meet y = x² + 1 at two points?

x² + 1 = mx − 3
x² − mx + 4 = 0
Discriminant: m² − 16 = (m − 4)(m + 4) > 0
m < −4 or m > 4

Solve a quadratic inequality like this with a quick sketch of the U-shaped graph. The discriminant is above the axis outside the roots.

Method for combined problems

Longer questions chain the skills together. Work in this order and label each result so that a slip in one part does not cost you the rest.

1. Draw a rough sketch with the given points. It catches sign errors early.
2. Find gradients first. Everything else (parallel, perpendicular, bisector) follows from them.
3. Write each line in y − y₁ = m(x − x₁) form, then convert to the requested form only at the end.
4. When two lines meet, solve them simultaneously and give both coordinates.
5. When a line meets a curve, substitute, then read the discriminant if the question mentions tangent, touching, distinct points or no intersection.
6. Substitute your answers back into both equations as a final check.

Presenting answers

• Give exact values: fractions such as 3/4 and surds such as 2√13, not rounded decimals, unless the question asks for a decimal.
• “Show that” questions need every line of working to be visible, finishing on the stated result.
• When a question gives the required form, such as ax + by + c = 0 with whole numbers, your final line must be in that form.
• State the conclusion in words for interpretation questions: “equal roots, so the line is a tangent”.

Must-know distinctions

• Parallel versus perpendicular. Parallel: same gradient. Perpendicular: flip and change sign.
• Gradient versus intercept in ax + by + c = 0. The gradient is −a/b. The intercept is −c/b.
• Distinct, equal, no real roots. These describe the quadratic after substitution, not the curve itself.
• Point of intersection versus point of contact. Contact is the single intersection point when the line is a tangent.
• Horizontal and vertical lines. y = k is horizontal with gradient 0. x = k is vertical with no gradient. They are perpendicular, but the product rule does not apply.
• Mid-point versus distance. The mid-point uses addition and halving. The distance uses subtraction, squaring, adding and a square root.

Quick self-test

1. Find the gradient of the line through (−2, 4) and (6, −8).
2. Find the mid-point of (−5, 2) and (9, 10).
3. Find the distance between (1, −2) and (13, 3).
4. State the gradient and y-intercept of 8x + 5y − 20 = 0.
5. Find the line through (4, 1) with gradient −2, in the form ax + by + c = 0.
6. What is the gradient of a line perpendicular to one with gradient −5/3?
7. Are 3x − 2y = 8 and 4x + 6y = 5 perpendicular?
8. Find the line through (2, 5) parallel to y = 4 − x, in the form x + y + c = 0.
9. For which values of k does y = x + k miss the curve y = x² − 3x + 6?
10. Find the points where y = 20/x meets 2x + y = 13.

Answers

1. −12/8 = −3/2.
2. (2, 6).
3. √(12² + 5²) = √169 = 13.
4. Gradient −8/5, intercept 4.
5. y − 1 = −2(x − 4), giving 2x + y − 9 = 0.
6. 3/5.
7. The gradients are 3/2 and −2/3, with product −1, so yes.
8. The gradient is −1. y − 5 = −(x − 2), so x + y − 7 = 0.
9. The discriminant is 4k − 8, which is negative when k < 2.
10. (5/2, 8) and (4, 5), as in the worked reminder above.

Where marks are usually lost

• Subtracting coordinates in different orders in the numerator and denominator of the gradient.
• Dropping the minus sign when reading the gradient from ax + by + c = 0.
• Taking the perpendicular gradient as only the reciprocal, or only the negative.
• Starting a perpendicular bisector from a point on AB other than the mid-point.
• Stopping at the x-values and not giving the y-coordinates of the intersection points.
• Substituting into the wrong equation and ending up with a messy expression instead of a quadratic.
• Setting the discriminant greater than zero when the question says tangent, or equal to zero when it says two points.
• Dropping a bracket: (2 − m)² written as 4 − m².
• Leaving fractions in a line equation when the question asks for whole-number coefficients.
• Giving a decimal approximation when an exact answer or surd was expected.

Official syllabus

These notes follow 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.