Coordinate geometry

Coordinate geometry is geometry done with coordinates: points are pairs (x, y), and lines and circles are equations. With a handful of formulas you can find the gradient, midpoint and length of a line segment, write the equation of a straight line, and find the centre and radius of a circle. It starts at GCSE and carries on into A-level, where the topic is called coordinate geometry in the (x, y) plane.

Coordinate geometry formulas

In Pearson Edexcel's GCSE, Higher tier students use y = mx + c to spot parallel and perpendicular lines, find the line through two points, and use the equation of a circle with centre at the origin, including its tangents. Edexcel's A-level (topic 3) adds y − y₁ = m(x − x₁), ax + by + c = 0, circles with any centre, and parametric equations.

Gradient, midpoint and distance

Take A(−2, 3) and B(4, 11). Going from A to B, x goes up by 4 − (−2) = 6 and y goes up by 11 − 3 = 8. Every one of the three formulas uses those two changes.

• Gradient: 8 ÷ 6 = 4/3. Up 4 for every 3 across.
• Midpoint: ((−2 + 4) ÷ 2, (3 + 11) ÷ 2) = (1, 7). Average the x values and the y values.
• Distance: √(6² + 8²) = √100 = 10. This is Pythagoras on the triangle in the diagram.

The Midpoint and Distance Calculator finds all three for any two points, and the distance formula and midpoint guides go into more detail.

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Midpoint and Distance Calculator

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Equation of a straight line

To find a line's equation you need its gradient and one point on it. The quickest form is the point slope form, y − y₁ = m(x − x₁). For the line through A(−2, 3) with gradient 4/3:

Multiply by 3 to clear the fraction: 3y − 9 = 4x + 8. Rearranged into the form ax + by + c = 0 with whole numbers, that is 4x − 3y + 17 = 0. Check it with B(4, 11): 16 − 33 + 17 = 0, so B is on the line.

Parallel and perpendicular lines

Parallel lines have the same gradient. Perpendicular lines have gradients that multiply to −1: flip the fraction and change the sign. A line perpendicular to AB has gradient −3/4.

Example: the perpendicular bisector of AB goes through the midpoint M(1, 7) with gradient −3/4. So y − 7 = −3/4 (x − 1), which gives 4y − 28 = −3x + 3, or 3x + 4y − 31 = 0. Every point on it is the same distance from A and B.

Where two lines meet

The crossing point satisfies both equations, so solve them as simultaneous equations. For y = 4x − 7 and x + 2y = 13, substitute: x + 2(4x − 7) = 13, so 9x − 14 = 13, 9x = 27 and x = 3. Then y = 4(3) − 7 = 5. The lines meet at (3, 5).

Try a pair yourself, then press Mark to check it in the algebra course.

Equation of a circle

Every point (x, y) on a circle is r away from the centre (a, b). Writing that with the distance formula and squaring both sides gives the equation of a circle:

With the centre at the origin it becomes x² + y² = r². Watch the signs: (x + 4)² means x − (−4), so the centre has x-coordinate −4.

Finding the centre and radius by completing the square

When the equation is expanded, such as x² + y² + 8x − 2y − 8 = 0, complete the square on the x terms and the y terms separately:

So the centre is (−4, 1) and the radius is √25 = 5. More methods are in how to find the centre of a circle.

The tangent to a circle

P(−1, 5) is on the circle, since (−1 + 4)² + (5 − 1)² = 9 + 16 = 25. A tangent meets the radius at 90°, so use the perpendicular gradient:

1. Gradient of the radius CP = (5 − 1) ÷ (−1 − (−4)) = 4/3.
2. Gradient of the tangent = −3/4.
3. Tangent: y − 5 = −3/4 (x + 1), so 4y − 20 = −3x − 3, which is 3x + 4y − 17 = 0.

Where a line meets a circle

Substitute the line into the circle and solve the quadratic. Does y = x + 10 meet the circle above? Substituting gives (x + 4)² + (x + 9)² = 25, which simplifies to x² + 13x + 36 = 0, or (x + 4)(x + 9) = 0. So x = −4 or x = −9, and the line crosses the circle at (−4, 6) and (−9, 1).

The discriminant b² − 4ac tells you how many points there are without solving: here 13² − 4 × 36 = 25, which is positive, so two points. Zero means the line is a tangent and touches once; negative means it misses the circle.

Circle facts you can use

Edexcel's A-level specification lists three circle theorems to use with coordinates:

• Angles in a semicircle are 90°. If AB is a diameter and P is on the circle, AP and BP are perpendicular, so their gradients multiply to −1.
• A chord is cut in half by the perpendicular to it from the centre. So the perpendicular bisector of any chord passes through the centre. Two of them cross at the centre, which is how you find the circle through three points.
• The radius at a point is perpendicular to the tangent there. That is the tangent method above.

Common mistakes

• Mixing the order in the gradient. Subtract both coordinates in the same order: (y₂ − y₁) over (x₂ − x₁).
• Dropping a negative. Subtracting a negative coordinate adds: 4 − (−2) = 6.
• Wrong signs for the centre. (x − 3)² means the centre is at x = 3, and (x + 3)² means x = −3.
• Giving r² as the radius. In (x + 4)² + (y − 1)² = 25, the radius is 5, not 25.
• Only changing the sign for a perpendicular gradient. Flip the fraction as well: the perpendicular to 4/3 is −3/4, not −4/3.
• Leaving fractions in ax + by + c = 0. Questions usually want a, b and c as integers, so multiply through.

Practise lines and circles under time pressure with today's A-level Maths: Algebra & Functions short timed paper. For the graphs side of the topic, try graphing linear functions.

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