Welcome to Coordinate Geometry in the (x, y) Plane
Welcome to one of the most visual and rewarding topics in AS 1: Pure Mathematics! Coordinate geometry bridges the gap between algebra and geometry. By placing geometric shapes onto a grid (the Cartesian plane), we can use algebra to solve geometric puzzles, find distances, calculate angles, and track intersections.
Why is this important? Every time you use GPS navigation on your smartphone, play a 3D video game, or look at digital animations, computer algorithms are using coordinate geometry behind the scenes to pinpoint locations and map trajectories!
Don't worry if algebra or geometry has felt intimidating in the past. We will break every single idea down into small, easy-to-follow steps with plenty of visual logic and practical tips.
---1. Line Segments: Distance, Midpoint, and Gradient
Let's start with two points on a plane: \(A(x_1, y_1)\) and \(B(x_2, y_2)\). A straight line segment connects them.
A. The Distance Between Two Points
Concept: The distance formula is simply Pythagoras' Theorem in disguise! The horizontal distance is \((x_2 - x_1)\) and the vertical distance is \((y_2 - y_1)\). The straight-line distance \(d\) is the hypotenuse.
The Formula:
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Example: Find the distance between \(A(1, 3)\) and \(B(4, 7)\).
1. Horizontal change: \(x_2 - x_1 = 4 - 1 = 3\)
2. Vertical change: \(y_2 - y_1 = 7 - 3 = 4\)
3. Apply formula: \(d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
B. The Midpoint of a Line Segment
Concept: The midpoint \(M\) is the exact center of the segment. To find it, you simply find the average of the \(x\)-coordinates and the average of the \(y\)-coordinates.
The Formula:
\(M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Memory Aid: Think "Add them up and divide by 2", just like finding the average test score of two exams!
C. Gradient (Slope) of a Line
Concept: Gradient, denoted by \(m\), measures the steepness and direction of a line. It is the ratio of vertical change ("rise") to horizontal change ("run").
The Formula:
\(m = \frac{y_2 - y_1}{x_2 - x_1}\)
Gradient Rules of Thumb:
• Sloping upwards from left to right: Positive gradient (\(m > 0\))
• Sloping downwards from left to right: Negative gradient (\(m < 0\))
• Perfectly horizontal line: Zero gradient (\(m = 0\))
• Perfectly vertical line: Undefined gradient (division by zero)
Key Takeaway: When working with distance, midpoint, or gradient, always label your points clearly as \((x_1, y_1)\) and \((x_2, y_2)\) before substituting them into formulas. Watch out for double negatives, e.g., \(3 - (-2) = 3 + 2 = 5\)!
---2. Equations of Straight Lines
A straight line equation describes the relationship between the \(x\) and \(y\) coordinates for every single point that lies on that line.
A. Standard Forms of a Straight Line
1. Slope-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).
2. Point-Gradient Form (The Exam Favourite!):
\(y - y_1 = m(x - x_1)\)
This form is the quickest and safest way to write the equation of a line when you know the gradient \(m\) and any point \((x_1, y_1)\) on the line.
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 \(a, b, c \in \mathbb{Z}\)."
B. Step-by-Step Example
Problem: Find the equation of the line passing through \((2, -3)\) and \((6, 5)\). Give 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 the point-gradient formula with the point \((2, -3)\).
\(y - (-3) = 2(x - 2)\)
\(y + 3 = 2x - 4\)
Step 3: Rearrange into the general form \(ax + by + c = 0\).
\(2x - y - 7 = 0\)
C. Parallel and Perpendicular Lines
Parallel Lines: Lines that never meet. They have identical steepness.
\(m_1 = m_2\)
Perpendicular Lines: Lines that intersect at right angles (\(90^\circ\)).
\(m_1 \times m_2 = -1\) or \(m_2 = -\frac{1}{m_1}\)
Rule of thumb: To find the perpendicular gradient, take the negative reciprocal (flip the fraction and change the sign). For example, if \(m = \frac{2}{3}\), the perpendicular gradient is \(-\frac{3}{2}\). If \(m = -4\), the perpendicular gradient is \(\frac{1}{4}\).
Key Takeaway: Whenever you see the word "perpendicular", immediately think: "Flip the fraction, flip the sign!"
---3. Coordinate Geometry of the Circle
A circle is defined as the set of all points that are an equal distance (the radius, \(r\)) from a fixed central point \((a, b)\).
A. Standard Equation of a Circle
Using Pythagoras' theorem once again, any point \((x, y)\) on the edge of a circle with center \((a, b)\) and radius \(r\) satisfies:
\((x - a)^2 + (y - b)^2 = r^2\)
Common Mistake to Avoid: Be careful with the signs of the center! If the equation is \((x - 3)^2 + (y + 5)^2 = 16\):
• Center is \((3, -5)\) (notice the sign flips: \(-3 \to +3\), \(+5 \to -5\))
• Radius is \(r = \sqrt{16} = 4\) (don't forget to take the square root of the right-hand side!)
B. Expanded / General Equation of a Circle
Sometimes you are given the circle equation in expanded form:
\(x^2 + y^2 + 2gx + 2fy + c = 0\)
For this form:
• Center: \((-g, -f)\)
• Radius: \(r = \sqrt{g^2 + f^2 - c}\)
How to find the center and radius by Completing the Square:
Example: Find the center and radius of \(x^2 + y^2 - 6x + 8y - 11 = 0\).
Step 1: Group \(x\) terms and \(y\) terms together, and move the constant to the other side.
\((x^2 - 6x) + (y^2 + 8y) = 11\)
Step 2: Complete the square for both brackets.
\((x - 3)^2 - 9 + (y + 4)^2 - 16 = 11\)
Step 3: Collect all numbers on the right-hand side.
\((x - 3)^2 + (y + 4)^2 = 11 + 9 + 16\)
\((x - 3)^2 + (y + 4)^2 = 36\)
Step 4: Read off the results.
• Center \(= (3, -4)\)
• Radius \(= \sqrt{36} = 6\)
Key Takeaway: Completing the square is the most reliable method to convert an expanded circle equation into the standard \((x - a)^2 + (y - b)^2 = r^2\) form.
---4. Intersections, Tangents, and Normals
A. Line and Circle Intersections
To find where a straight line and a circle intersect, solve their equations simultaneously by substituting the linear equation into the circle equation. This produces a quadratic equation in one variable.
Recall the discriminant (\(b^2 - 4ac\)) of the resulting quadratic:
• \(b^2 - 4ac > 0\): 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 (a tangent).
• \(b^2 - 4ac < 0\): The line does not intersect the circle.
B. Tangents and Normals to a Circle
• A tangent is a straight line that touches a circle at exactly one point \(P\).
• A normal is a straight line that is perpendicular to the tangent at \(P\). Crucial Fact: The normal always passes straight through the center of the circle!
Key Geometric Property: The radius to the point of contact \(P\) is perpendicular to the tangent line at \(P\).
Step-by-Step: Finding the Equation of a Tangent at a Point \(P(x_1, y_1)\):
1. Find the center of the circle, \(C(a, b)\).
2. Find the gradient of the radius \(CP\):
\(m_{\text{radius}} = \frac{y_1 - b}{x_1 - a}\)
3. Find the gradient of the tangent using perpendicular gradients:
\(m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}}\)
4. Write the tangent's equation using the point \(P(x_1, y_1)\):
\(y - y_1 = m_{\text{tangent}}(x - x_1)\)
C. Important Circle Theorems in Coordinate Geometry
Exam questions often combine coordinate geometry with basic circle theorems:
1. Angle in a Semicircle: If \(A\) and \(B\) are endpoints of a diameter, and \(P\) is any point on the circumference, the angle \(\angle APB = 90^\circ\). Therefore, gradient \(AP \times\) gradient \(PB = -1\).
2. Perpendicular Bisector of a Chord: The perpendicular bisector of any chord passes directly through the center of the circle.
3. Distance from Center to Tangent: The perpendicular distance from the center of the circle to any tangent line is exactly equal to the radius \(r\).
Key Takeaway: You don't need calculus to find tangents to circles! Always use the radius gradient and take the perpendicular negative reciprocal.
---5. Quick Review & Formula Cheat Sheet
Straight Lines:
• Gradient: \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
• Distance: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
• Midpoint: \(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)
• Point-Gradient Line Form: \(y - y_1 = m(x - x_1)\)
• Parallel Condition: \(m_1 = m_2\)
• Perpendicular Condition: \(m_1 \times m_2 = -1 \implies m_2 = -\frac{1}{m_1}\)
Circles:
• Standard Form: \((x - a)^2 + (y - b)^2 = r^2\) \(\implies\) Center \((a, b)\), Radius \(r\)
• Expanded Form: \(x^2 + y^2 + 2gx + 2fy + c = 0\) \(\implies\) Center \((-g, -f)\), Radius \(\sqrt{g^2 + f^2 - c}\)
• Tangent is perpendicular to the radius: \(m_{\text{tangent}} \times m_{\text{radius}} = -1\)
Exam Pro-Tip: Always draw a quick sketch of the line or circle! A 10-second rough diagram will help you check if your gradient should be positive or negative and ensure your center coordinates make visual sense.