Mathematics (9660) | Coordinate geometry in the (x, y) plane

👋 Welcome to Coordinate Geometry!

Hello! This chapter, P1.2 and P2.3, is all about mastering the maths of shapes and movement on the familiar \((x, y)\) plane. Coordinate geometry links algebra (equations) with geometry (pictures), making it one of the most useful and foundational topics in Pure Mathematics.

Whether you are calculating the distance between two cities or figuring out the perfect angle for a satellite dish, the principles you learn here are essential. Don't worry if coordinates sometimes feel like a treasure map; we’ll break down every tool you need to navigate the plane successfully!

🗺️ P1.2: Essential Tools for Straight Lines

1. The Distance Between Two Points

How do you find the length of a segment connecting point A to point B? We use the Distance Formula, which is really just an application of Pythagoras’ theorem!

If you have two points, \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the distance \(D\) is:

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

Analogy: Imagine driving along a grid. You travel the horizontal change (\(x\)-difference) and the vertical change (\(y\)-difference). The distance formula finds the shortest, straight-line path (the hypotenuse).

2. Midpoint Formula

The midpoint is the exact center point between two coordinates. To find it, you simply find the average of the \(x\)-coordinates and the average of the \(y\)-coordinates.

The midpoint \(M\) of \(A(x_1, y_1)\) and \(B(x_2, y_2)\) is:

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

This calculation is always straightforward, but make sure you don't accidentally subtract instead of add!

3. The Gradient (Slope) of a Line

The gradient (\(m\)) tells us how steep a line is. It is the ratio of the vertical change (rise) to the horizontal change (run).

$$m = \frac{\text{Change in } y}{\text{Change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$$

Key Takeaway:

• A positive gradient means the line goes uphill (from left to right).
• A negative gradient means the line goes downhill.
• A gradient of zero means it's a horizontal line (\(y = c\)).
• An undefined gradient means it's a vertical line (\(x = c\)).

4. Equations of Straight Lines

You must be comfortable with the three main forms of a straight-line equation:

i. Gradient-Intercept Form

$$y = mx + c$$ Where \(m\) is the gradient and \(c\) is the y-intercept (where the line crosses the \(y\)-axis). This is the most common and useful form for plotting.

ii. Point-Gradient Form

$$y - y_1 = m(x - x_1)$$ This is often the easiest form to use when you know the gradient \(m\) and just one point \((x_1, y_1)\). You use this to construct the equation before rearranging it into other forms.

iii. General Form

$$ax + by + c = 0$$ This form (where \(a, b, c\) are integers, and \(a\) is usually positive) is required for some exam questions, particularly those involving perpendicular distance (though perpendicular distance formulas themselves are usually outside the AS scope, recognizing this form is essential).

5. Parallel and Perpendicular Lines

This is a critical concept for geometry problems.

Parallel Lines

Two lines are parallel if they have the same steepness.

Condition: The gradients must be equal. $$m_1 = m_2$$

Perpendicular Lines

Two lines are perpendicular (meet at a 90° angle) if their gradients are negative reciprocals of each other.

Condition: The product of their gradients must be \(-1\). $$m_1 m_2 = -1 \quad \text{or} \quad m_2 = -\frac{1}{m_1}$$

If the gradient of the first line is \(m_1 = \frac{2}{3}\), the perpendicular gradient is \(m_2 = -\frac{3}{2}\). Flip it and negate it!

🛑 Common Mistake to Avoid:
When finding the gradient perpendicular to \(m=4\), students sometimes write \(m_{\perp} = -\frac{1}{4}\). Remember that 4 is \(\frac{4}{1}\). Always flip and negate!

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🔗 P1.2: Intersection of Lines and Curves

Finding the point(s) where two graphs cross is a core skill. Geometrically, the intersection represents the coordinates \((x, y)\) that satisfy both equations simultaneously.

1. Intersection of Two Straight Lines

This is solved simply by using simultaneous equations. If you have:
1. \(y = 2x + 1\)
2. \(y = -x + 4\)
You can equate them (\(2x + 1 = -x + 4\)) to find \(x\), and then substitute \(x\) back into either equation to find \(y\).

2. Intersection of a Straight Line and a Curve

The syllabus requires you to solve the intersection of one linear equation (the straight line) and one quadratic equation (the curve).

Step-by-Step Process:

1. Isolate \(y\): Ensure the linear equation is in the form \(y = \ldots\)
2. Substitute: Substitute the expression for \(y\) from the linear equation into the quadratic curve equation.
3. Solve the Quadratic: This will create a single quadratic equation in terms of \(x\) (e.g., \(Ax^2 + Bx + C = 0\)). Solve this equation using factorisation or the quadratic formula.
4. Find \(y\): Substitute the value(s) of \(x\) you found back into the linear equation to find the corresponding \(y\) coordinates.

3. Geometrical Interpretation using the Discriminant

The key insight here is how the discriminant (\(\Delta = b^2 - 4ac\)) of the resulting quadratic equation relates to the geometry of the intersection.

When you solve the combined quadratic equation \(Ax^2 + Bx + C = 0\):

• If \(\mathbf{b^2 - 4ac > 0}\): The equation has two distinct real roots. This means the line intersects the curve at two distinct points.
• If \(\mathbf{b^2 - 4ac = 0}\): The equation has one repeated real root. This means the line is a tangent to the curve, touching it at exactly one point.
• If \(\mathbf{b^2 - 4ac < 0}\): The equation has no real roots. This means the line and the curve do not intersect at all.

This relationship is often tested when a question asks you to prove that a line is tangent to a curve (set \(b^2-4ac = 0\)) or find the range of values for which the line misses the curve (set \(b^2-4ac < 0\)).

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🚀 P2.3: Advanced Coordinate Geometry (Parametrics)

In AS Pure Maths (P1.2), we mainly dealt with Cartesian equations (like \(y = x^2\)). In P2.3, we introduce a new, powerful way to describe curves: Parametric Equations.

1. What are Parametric Equations?

Instead of relating \(x\) and \(y\) directly, parametric equations introduce a third variable, called the parameter (usually \(t\) or \(\theta\)).

The coordinates of the curve are given by two separate functions of this parameter:
$$x = f(t) \quad \text{and} \quad y = g(t)$$

Analogy: Think of \(t\) as time. As time passes, the \(x\) and \(y\) coordinates change, tracing out the path of the curve. This is especially useful for describing motion or complex shapes like cycloids.

Example from the Syllabus:
Algebraic parameter: \(x = t^2\), \(y = 2t\)
Trigonometric parameter: \(x = a \cos \theta\), \(y = b \sin \theta\) (This describes an ellipse, or a circle if \(a=b\))

2. Conversion: Parametric to Cartesian

The main skill here is to eliminate the parameter (\(t\) or \(\theta\)) to get the standard Cartesian form \(y = F(x)\).

Case 1: Algebraic Parameters (e.g., using \(t\))

Goal: Make \(t\) the subject of one equation, and substitute it into the other.

Example: Convert \(x = t^2\) and \(y = 2t\) to Cartesian form.

1. From the second equation: \(t = \frac{y}{2}\).
2. Substitute this expression for \(t\) into the first equation: $$x = \left(\frac{y}{2}\right)^2$$
3. Simplify to get the Cartesian form: \(x = \frac{y^2}{4}\) or \(\mathbf{y^2 = 4x}\) (a parabola).

Case 2: Trigonometric Parameters (e.g., using \(\theta\))

Goal: Use the identity \(\mathbf{\sin^2 \theta + \cos^2 \theta = 1}\).

Example: Convert \(x = 5 \cos \theta\) and \(y = 5 \sin \theta\) to Cartesian form.

1. Isolate the trigonometric functions: $$\cos \theta = \frac{x}{5} \quad \text{and} \quad \sin \theta = \frac{y}{5}$$
2. Square both and substitute into the identity: $$\left(\frac{x}{5}\right)^2 + \left(\frac{y}{5}\right)^2 = 1$$
3. Simplify: \(\frac{x^2}{25} + \frac{y^2}{25} = 1\), which gives the Cartesian form: \(\mathbf{x^2 + y^2 = 25}\) (a circle).

Tips for Success: When dealing with parametrically defined curves, remember that all the skills from Differentiation (P1.3) and Integration (P1.4) can also be applied using the chain rule to find \(\frac{dy}{dx}\) or the area under the curve, although specific application of calculus to parametrics is usually reserved for P2.6/P2.7. For P2.3, the focus is purely on understanding and converting the equations themselves.

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