Co-ordinate geometry in the (x, y) plane

Q: What is the formula for the gradient \(m\) of a line passing through \((x_1, y_1)\) and \((x_2, y_2)\)?

\(m = \frac{y_2 - y_1}{x_2 - x_1}\)

Q: What is the relationship between the gradients of two perpendicular lines?

\(m_1 \times m_2 = -1\) (or \(m_2 = -\frac{1}{m_1}\))

Q: What two conditions define the perpendicular bisector of a line segment?

It passes through the midpoint of the segment and has a gradient that is the negative reciprocal of the segment's gradient.

Q: How do you find the coordinates of the point of intersection of two straight lines?

Solve their linear equations simultaneously using substitution or elimination.

Q: What is the standard equation of a circle with centre \((a, b)\) and radius \(r\)?

\((x - a)^2 + (y - b)^2 = r^2\)

Q: How do you find the centre and radius from an expanded circle equation \(x^2 + y^2 + 2gx + 2fy + c = 0\)?

Complete the square for both the \(x\) terms and the \(y\) terms to rewrite it in the form \((x - a)^2 + (y - b)^2 = r^2\).

Q: What is the geometric relationship between a circle's radius and its tangent at the point of contact?

The tangent is perpendicular to the radius at the point of contact (\(m_r \times m_t = -1\)).

Mathematics 2210 · CCEA AS Level · 10 min read · Updated: 31 Aug 2026

Welcome to Co-ordinate Geometry in the \((x, y)\) Plane

Welcome to one of the most visual and powerful topics in AS Pure Mathematics! Co-ordinate geometry links algebra and geometry together. By plotting equations on a grid, algebraic expressions turn into lines, curves, and shapes.

Think of the Cartesian coordinate plane like a GPS navigation system: every point on a map has a unique address given by \((x, y)\). In this chapter, you will master straight lines and circles, learn how to find their properties, and solve problems involving where they meet. Don't worry if algebra has felt abstract in the past—coordinate geometry lets you literally see how the equations work!

1. The Fundamentals: Distance, Midpoint, and Gradient

Let us begin with two points on a grid: \(A(x_1, y_1)\) and \(B(x_2, y_2)\).

A. The Distance Between Two Points

The distance formula finds the length of the straight line segment joining \(A\) and \(B\). It is simply Pythagoras' Theorem in disguise!

Formula:

\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Real-World Analogy: Imagine walking across a rectangular park from the southwest corner to the northeast corner. Instead of walking east and then north, the distance formula calculates the direct diagonal path.

Quick Example: Find the distance between \(A(1, 4)\) and \(B(4, 8)\).
• Horizontal difference: \(x_2 - x_1 = 4 - 1 = 3\)
• Vertical difference: \(y_2 - y_1 = 8 - 4 = 4\)
• Distance: \(d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

B. The Midpoint of a Line Segment

The midpoint \(M\) is the point exactly halfway between \(A\) and \(B\). You simply find the average of the \(x\)-coordinates and the average of the \(y\)-coordinates.

Formula:

\(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)

Quick Example: The midpoint of \((2, -3)\) and \((8, 7)\) is:
\(M = \left(\frac{2 + 8}{2}, \frac{-3 + 7}{2}\right) = \left(\frac{10}{2}, \frac{4}{2}\right) = (5, 2)\)

C. The Gradient of a Straight Line

The gradient, denoted by \(m\), measures the steepness and direction of a line.

Formula:

\(m = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}\)

• Positive gradient (\(m > 0\)): Line slopes upwards from left to right.
• Negative gradient (\(m < 0\)): Line slopes downwards from left to right.
• Zero gradient (\(m = 0\)): Line is completely horizontal (e.g., \(y = 3\)).
• Undefined gradient: Line is completely vertical (e.g., \(x = -2\)).

Key Takeaway for Fundamentals:
• Distance = Pythagoras (\(\sqrt{\Delta x^2 + \Delta y^2}\))
• Midpoint = Average coordinates (\(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\))
• Gradient = Slope (\(\frac{\Delta y}{\Delta x}\))

2. Equations of a Straight Line

There are two main standard formats used at AS Level to write the equation of a straight line.

Format 1: Gradient-Intercept Form

\(y = mx + c\)

Here, \(m\) is the gradient and \(c\) is the \(y\)-intercept (the point \((0, c)\) where the line crosses the \(y\)-axis).

Format 2: Point-Gradient Form (The Most Useful Tool!)

\(y - y_1 = m(x - x_1)\)

Whenever you know the gradient \(m\) and any point \((x_1, y_1)\) on the line, use this formula directly. It avoids intermediate steps and reduces calculation errors.

Format 3: General Form

\(ax + by + c = 0\)

where \(a\), \(b\), and \(c\) are integers. Exam questions often ask: "Give your answer in the form \(ax + by + c = 0\) where \(a, b, c \in \mathbb{Z}\)."

Step-by-Step Example:
Find the equation of the line passing through \((2, -3)\) and \((6, 5)\), giving your answer in the form \(ax + by + c = 0\).

Step 1: Find the gradient \(m\).
\(m = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2\)

Step 2: Use \(y - y_1 = m(x - x_1)\) with the point \((2, -3)\).
\(y - (-3) = 2(x - 2)\)
\(y + 3 = 2x - 4\)

Step 3: Rearrange into the form \(ax + by + c = 0\).
\(2x - y - 7 = 0\)

Key Takeaway for Line Equations: Whenever you need the equation of a line, your mission is always to find two things: one point and the gradient.

3. Parallel and Perpendicular Lines

A. Parallel Lines

Parallel lines travel in the exact same direction and never cross. They have equal gradients.

If Line 1 has gradient \(m_1\) and Line 2 has gradient \(m_2\), then:
\(m_1 = m_2\)

B. Perpendicular Lines

Perpendicular lines meet at right angles (\(90^\circ\)). Their gradients are the negative reciprocals of each other.

Rule:
\(m_1 \times m_2 = -1\) or \(m_2 = -\frac{1}{m_1}\)

Simple Memory Trick: To find a perpendicular gradient, flip the fraction upside down and change the sign.

• If \(m = 3\), the perpendicular gradient is \(-\frac{1}{3}\).
• If \(m = -\frac{2}{5}\), the perpendicular gradient is \(+\frac{5}{2}\).

C. Perpendicular Bisectors

A perpendicular bisector of a line segment \(AB\) is a line that does two things:
1. Cuts the segment \(AB\) in half (passes through the midpoint of \(AB\)).
2. Crosses \(AB\) at a right angle (its gradient is \(-\frac{1}{m_{AB}}\)).

Method to find a perpendicular bisector:
1. Find the midpoint of \(AB\).
2. Find the gradient of \(AB\).
3. Take the negative reciprocal of that gradient.
4. Use \(y - y_1 = m_{\perp}(x - x_1)\) with the midpoint.

Key Takeaway:
• Parallel: \(m_1 = m_2\)
• Perpendicular: \(m_1 m_2 = -1\)
• Perpendicular Bisector: Midpoint + Perpendicular Gradient

4. Intersections of Straight Lines

To find the point where two straight lines intersect, solve their equations simultaneously using either elimination or substitution.

Worked Example:
Find the coordinates of the point where \(2x + y = 7\) and \(3x - 2y = 7\) cross.

Step 1: Rearrange the first equation to express \(y\) in terms of \(x\):
\(y = 7 - 2x\)

Step 2: Substitute this expression into the second equation:
\(3x - 2(7 - 2x) = 7\)
\(3x - 14 + 4x = 7\)
\(7x - 14 = 7\)
\(7x = 21 \implies x = 3\)

Step 3: Substitute \(x = 3\) back to find \(y\):
\(y = 7 - 2(3) = 1\)

The point of intersection is \((3, 1)\).

5. The Coordinate Geometry of Circles

A. What is a Circle?

A circle is the set of all points that are a fixed distance (the radius, \(r\)) from a fixed point (the centre, \((a, b)\)).

B. Standard Equation of a Circle

Formula:

\((x - a)^2 + (y - b)^2 = r^2\)

• Centre = \((a, b)\)
• Radius = \(r = \sqrt{r^2}\)

Watch out for the signs!
• \((x - 3)^2 + (y - 5)^2 = 16\) has centre \((3, 5)\) and radius \(\sqrt{16} = 4\).
• \((x + 2)^2 + (y - 4)^2 = 25\) can be rewritten as \((x - (-2))^2 + (y - 4)^2 = 5^2\), so its centre is \((-2, 4)\) and radius is \(5\).

C. Expanded Form and Completing the Square

Sometimes the equation of a circle is given in expanded form:
\(x^2 + y^2 + 2gx + 2fy + c = 0\)

To find the centre and radius from this form, complete the square for the \(x\) terms and the \(y\) terms separately.

Step-by-Step Example:
Find the centre and radius of the circle with equation \(x^2 + y^2 - 6x + 8y - 11 = 0\).

Step 1: Group the \(x\) and \(y\) terms together:
\((x^2 - 6x) + (y^2 + 8y) = 11\)

Step 2: Complete the square for both brackets:
\((x - 3)^2 - 9 + (y + 4)^2 - 16 = 11\)

Step 3: Move the constant numbers to the right-hand side:
\((x - 3)^2 + (y + 4)^2 = 11 + 9 + 16\)
\((x - 3)^2 + (y + 4)^2 = 36\)

Conclusion: Centre is \((3, -4)\) and radius is \(r = \sqrt{36} = 6\).

Key Takeaway for Circle Equations:
Always rearrange into \((x - a)^2 + (y - b)^2 = r^2\). Remember to take the square root of the right-hand side to get the radius!

6. Geometric Properties of Circles

Coordinate geometry problems often use classic circle theorems from GCSE. Here are the most essential rules you must know:

Rule 1: Tangent is Perpendicular to the Radius

A tangent touches a circle at exactly one point \(P\). The radius drawn to \(P\) meets the tangent at right angles (\(90^\circ\)).

• \(\text{Gradient of radius } (m_r) \times \text{Gradient of tangent } (m_t) = -1\)
• To find the equation of a tangent at point \(P\):
1. Find the centre of the circle \(C\).
2. Find the gradient of the radius \(CP\).
3. Find the perpendicular gradient: \(m_t = -\frac{1}{m_{CP}}\).
4. Use \(y - y_1 = m_t(x - x_1)\) with the point \(P\).

Rule 2: The Normal to a Circle

The normal to a circle at point \(P\) is the line perpendicular to the tangent at \(P\). This means the normal always passes through the centre of the circle!

Rule 3: Angle in a Semicircle is a Right Angle

If \(A\) and \(B\) are the endpoints of a diameter, and \(P\) is any point on the circumference, then the angle \(\angle APB = 90^\circ\). This means the line segment \(AP\) is perpendicular to \(BP\), so \(m_{AP} \times m_{BP} = -1\).

Rule 4: Perpendicular Bisector of a Chord

A chord is any line segment connecting two points on a circle. The perpendicular bisector of any chord always passes straight through the centre of the circle.

Did you know? If you are given three points on a circle, you can find the centre by finding the equations of the perpendicular bisectors of two chords and finding where they intersect!

7. Intersections of Lines and Circles

To find where a straight line meets a circle, substitute the linear equation into the circle equation. This will produce a quadratic equation.

Recall the discriminant \(\Delta = b^2 - 4ac\):
• \(b^2 - 4ac > 0\): The line crosses the circle at two distinct points (a secant line).
• \(b^2 - 4ac = 0\): The line touches the circle at exactly one point (the line is a tangent).
• \(b^2 - 4ac < 0\): The line does not intersect the circle.

Worked Example:
Show that the line \(y = x - 5\) is a tangent to the circle \(x^2 + y^2 = 12.5\) and find the point of contact.

Step 1: Substitute \(y = x - 5\) into the circle equation:
\(x^2 + (x - 5)^2 = 12.5\)
\(x^2 + x^2 - 10x + 25 = 12.5\)
\(2x^2 - 10x + 12.5 = 0\)

Step 2: Multiply by \(2\) to clear decimals:
\(4x^2 - 20x + 25 = 0\)

Step 3: Check the discriminant \(b^2 - 4ac\):
\((-20)^2 - 4(4)(25) = 400 - 400 = 0\)
Since the discriminant equals \(0\), there is exactly one repeated root. Therefore, the line is a tangent.

Step 4: Factorise to find the point of contact:
\((2x - 5)^2 = 0 \implies 2x - 5 = 0 \implies x = 2.5\)
Find \(y\): \(y = 2.5 - 5 = -2.5\)
Point of contact is \((2.5, -2.5)\).

8. Quick Revision Checklist & Common Mistakes to Avoid

Common Mistakes:
• Sign errors with coordinates: When using \((x - a)^2 + (y - b)^2 = r^2\), remember that \((x + 4)\) means \(a = -4\).
• Forgetting to square root \(r^2\): If the equation equals \(49\), the radius is \(7\), not \(49\).
• Gradient formula flips: Always calculate \(\frac{\Delta y}{\Delta x}\) (vertical over horizontal), never \(\frac{\Delta x}{\Delta y}\).
• Negative reciprocal slips: The perpendicular to \(-\frac{1}{4}\) is \(+4\), not \(-\frac{1}{4}\) or \(-4\).

Summary Formula Card:
• Gradient: \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
• Line Equation: \(y - y_1 = m(x - x_1)\)
• Perpendicular Lines: \(m_1 m_2 = -1\)
• Distance: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
• Midpoint: \(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)
• Circle: \((x - a)^2 + (y - b)^2 = r^2\) with centre \((a, b)\) and radius \(r\)

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the formula to calculate the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\)?

\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

How do you find the midpoint of a line segment connecting \((x_1, y_1)\) and \((x_2, y_2)\)?

\(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)

What is the formula for the gradient \(m\) of a line passing through \((x_1, y_1)\) and \((x_2, y_2)\)?

\(m = \frac{y_2 - y_1}{x_2 - x_1}\)

What is the point-gradient formula for the equation of a straight line?

\(y - y_1 = m(x - x_1)\)

What is the relationship between the gradients of two perpendicular lines?

\(m_1 \times m_2 = -1\) (or \(m_2 = -\frac{1}{m_1}\))

What two conditions define the perpendicular bisector of a line segment?

It passes through the midpoint of the segment and has a gradient that is the negative reciprocal of the segment's gradient.

How do you find the coordinates of the point of intersection of two straight lines?

Solve their linear equations simultaneously using substitution or elimination.

What is the standard equation of a circle with centre \((a, b)\) and radius \(r\)?

\((x - a)^2 + (y - b)^2 = r^2\)

How do you find the centre and radius from an expanded circle equation \(x^2 + y^2 + 2gx + 2fy + c = 0\)?

Complete the square for both the \(x\) terms and the \(y\) terms to rewrite it in the form \((x - a)^2 + (y - b)^2 = r^2\).

What is the geometric relationship between a circle's radius and its tangent at the point of contact?

The tangent is perpendicular to the radius at the point of contact (\(m_r \times m_t = -1\)).

Through which point does the normal to a circle at any point on its circumference always pass?

The centre of the circle.

How can the discriminant \(\Delta = b^2 - 4ac\) be used to determine how a straight line intersects a circle?

\(b^2 - 4ac > 0\): two intersection points (secant); \(b^2 - 4ac = 0\): one point (tangent); \(b^2 - 4ac < 0\): no intersection.

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