Further Pure Mathematics | Rectangular Cartesian coordinates

🗺️ Welcome to Rectangular Cartesian Coordinates!

Hello Future Mathematicians! This chapter, Rectangular Cartesian Coordinates, is your mathematical roadmap. It’s all about describing where things are and how far apart they are on a flat surface (the coordinate plane).

In Further Pure Mathematics, we don't just plot points; we use the coordinates to calculate crucial properties like lengths, slopes, and areas. Mastering these tools is fundamental for success in later chapters! Don't worry if geometry isn't your favourite—we’ll break everything down into manageable steps.

Key Terminology Review

• Origin: The point \((0, 0)\).
• Coordinates: A pair of numbers \((x, y)\) representing a position.
• Abscissa: The x-coordinate (horizontal position).
• Ordinate: The y-coordinate (vertical position).

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📐 Section 1: Measuring the Line Segment

When you have two points, \(A(x_1, y_1)\) and \(B(x_2, y_2)\), we often need to know the length of the line segment AB and exactly where the middle point is.

1.1 The Distance Formula (Length)

The distance formula is simply the application of the Pythagorean Theorem. If you draw a right-angled triangle using the line segment AB as the hypotenuse, the horizontal distance is \((x_2 - x_1)\) and the vertical distance is \((y_2 - y_1)\).

The Formula:

The distance \(d\) between \(A(x_1, y_1)\) and \(B(x_2, y_2)\) is: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Quick Tip: Because we are squaring the differences, it doesn't matter which point you choose as \((x_1, y_1)\) or \((x_2, y_2)\). The result will be the same!

Step-by-Step Example: Find the distance between \(A(1, 5)\) and \(B(4, 1)\).

1. Find the difference in \(x\): \(4 - 1 = 3\).
2. Find the difference in \(y\): \(1 - 5 = -4\).
3. Square and add: \(3^2 + (-4)^2 = 9 + 16 = 25\).
4. Take the square root: \(\sqrt{25} = 5\). The distance is 5 units.

1.2 The Midpoint Formula

Finding the midpoint is like finding the average position. You average the \(x\)-coordinates and average the \(y\)-coordinates separately.

The Formula:

The midpoint \(M\) of the line segment AB is: $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

Analogy: Imagine you are meeting a friend halfway. You need to travel half of the total horizontal distance and half of the total vertical distance. Averaging the coordinates tells you exactly where that halfway point is!

Key Takeaway for Section 1: Distance uses subtraction and Pythagoras (squaring). Midpoint uses addition and division by two (averaging).

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📈 Section 2: Understanding the Slope (Gradient)

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

2.1 Calculating the Gradient \(m\)

The gradient is calculated as the ratio of the change in \(y\) (vertical rise) to the change in \(x\) (horizontal run).

The Formula:

The gradient \(m\) between \(A(x_1, y_1)\) and \(B(x_2, y_2)\) is: $$m = \frac{\text{Change in } y}{\text{Change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$$

Interpreting the Gradient:

• Positive \(m\): The line goes up from left to right (uphill).
• Negative \(m\): The line goes down from left to right (downhill).
• \(m = 0\): The line is horizontal.
• Undefined \(m\): The line is vertical (denominator \(x_2 - x_1 = 0\)).

2.2 Parallel and Perpendicular Lines

The gradient is incredibly useful for determining the geometric relationship between two lines, \(L_1\) (with gradient \(m_1\)) and \(L_2\) (with gradient \(m_2\)).

1. Parallel Lines

Two lines are parallel if they have the same gradient. They never intersect.

$$m_1 = m_2$$

2. Perpendicular Lines

Two lines are perpendicular (meet at a 90° angle) if the product of their gradients is \(-1\).

$$m_1 \times m_2 = -1 \quad \text{or} \quad m_2 = -\frac{1}{m_1}$$

Memory Aid: To find the perpendicular gradient, you must flip it and change the sign (negative reciprocal).
Example: If \(m_1 = \frac{2}{3}\), then \(m_2 = -\frac{3}{2}\).

Did you know? This relationship is the key mathematical tool we use to find the equation of altitudes and perpendicular bisectors of geometric shapes.

Key Takeaway for Section 2: Gradient (\(m\)) is rise/run. Parallel lines have equal \(m\). Perpendicular lines have negative reciprocal \(m\).

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📝 Section 3: Equations of Straight Lines

The equation of a line provides a relationship between the \(x\) and \(y\) coordinates for every single point on that line. To define a straight line, you usually need two things: the gradient (\(m\)) and a point the line passes through.

3.1 The Slope-Intercept Form: \(y = mx + c\)

This is the most famous form, easily showing the steepness and where the line crosses the \(y\)-axis.

• \(m\) is the gradient.
• \(c\) is the \(y\)-intercept (the point \((0, c)\)).

Process: If you are given the gradient \(m\) and a point \((x_1, y_1)\), you can substitute these values into \(y = mx + c\) and solve for \(c\).

3.2 The Point-Gradient Form: \(y - y_1 = m(x - x_1)\)

For Further Pure Mathematics, this form is often faster and more powerful. It allows you to write the equation immediately if you know the gradient \(m\) and any point \((x_1, y_1)\) on the line.

Step-by-Step Example: Find the equation of the line with gradient \(m = -3\) passing through the point \((2, 5)\).

1. Identify \(m = -3\), \(x_1 = 2\), \(y_1 = 5\).
2. Substitute into the point-gradient form: \(y - 5 = -3(x - 2)\).
3. Simplify (rearrange into \(y = mx + c\) form if required):
   \(y - 5 = -3x + 6\)
   \(y = -3x + 11\)

Key Takeaway for Section 3: The Point-Gradient form, \(y - y_1 = m(x - x_1)\), is the most efficient starting point for solving line equations.

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🔍 Section 4: Applications – Intersections and Area

4.1 Finding the Point of Intersection

The point of intersection of two lines is the single coordinate \((x, y)\) that satisfies both equations simultaneously. Finding it requires solving the two equations as a pair of simultaneous equations.

Example: Find the intersection of \(L_1: y = 2x + 1\) and \(L_2: 3x + y = 6\).

Step 1: Use substitution (since \(y\) is isolated in \(L_1\)). Substitute \(2x + 1\) for \(y\) in \(L_2\).
$$3x + (2x + 1) = 6$$

Step 2: Solve for \(x\).
$$5x + 1 = 6$$ $$5x = 5 \implies x = 1$$

Step 3: Substitute \(x=1\) back into either original equation (use \(L_1\) as it’s simpler) to find \(y\).
$$y = 2(1) + 1 \implies y = 3$$

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

4.2 Area of Triangles and Quadrilaterals

In Further Pure Mathematics, you often encounter problems asking for the area of a polygon (like a triangle ABC or a quadrilateral ABCD) defined by its coordinates.

The most reliable method, especially when the shapes are not right-angled, is the Decomposition (Box) Method.

Process: Area Calculation (The Box Method)

1. Enclose the polygon (e.g., triangle ABC) completely within a large rectangle whose sides are parallel to the \(x\) and \(y\) axes.
2. Calculate the area of this large rectangle.
3. Calculate the areas of the right-angled triangles formed in the corners, outside the desired polygon.
4. Subtract the areas of the outer triangles from the area of the large rectangle to find the area of the polygon.

Why this method? Since the sides of the large box are horizontal and vertical, calculating the side lengths and areas of the outer right-angled triangles involves simple subtraction of coordinates, making it very straightforward and accurate.

Example for Triangle ABC:
If A=(2, 6), B=(8, 3), C=(4, 1).
The enclosing rectangle would go from \(x=2\) to \(x=8\) (width 6) and \(y=1\) to \(y=6\) (height 5).
Area of Rectangle = \(6 \times 5 = 30\).
You would then calculate the areas of the three corner triangles (using \( \frac{1}{2} \times \text{base} \times \text{height}\)) and subtract them from 30.

Key Takeaway for Section 4: Intersections are solved using simultaneous equations. Complex areas are most easily found by subtracting the areas of surrounding right-angled triangles from the area of an enclosing rectangle.