Mathematics - Additional (0606) | Straight-line graphs

📚 Additional Mathematics (0606) Study Notes: Straight-Line Graphs

Welcome to the chapter on Straight-Line Graphs! You might think this is simple material you covered in basic math, but in Additional Mathematics, we take these concepts much further. We will use the simple structure of a straight line to analyze complex, non-linear relationships. Mastering this chapter is essential for simplifying difficult equations and solving coordinate geometry problems efficiently. Let's dive in!

1. The Fundamentals of a Straight Line (Y = MX + C)

Every straight line on a Cartesian plane can be described by a linear equation. We rely heavily on the standard form you already know.

1.1 The Standard Equation

The core equation of a straight line is:

\[y = mx + c\]

• \(m\) is the Gradient (or slope). It tells us how steep the line is and its direction (positive or negative).
• \(c\) is the y-intercept. This is the point where the line crosses the y-axis (where \(x = 0\)).

Analogy: Think of a road. \(m\) is the steepness (how much you climb vertically for a given distance horizontally), and \(c\) is the starting height of the road.

1.2 Calculating the Gradient (\(m\))

If you have two points, \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the gradient is calculated as the change in \(y\) divided by the change in \(x\) (rise over run):

\[m = \frac{y_2 - y_1}{x_2 - x_1}\]

1.3 Finding the Equation of a Line

If you know the gradient \(m\) and one point \((x_1, y_1)\), the fastest way to find the equation is using the Point-Gradient Form:

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

Once you plug in the values, simply rearrange it into the standard \(y = mx + c\) form.

2. Parallel and Perpendicular Lines

A crucial skill in Additional Mathematics is using the gradient to determine the relationship between two lines.

2.1 Parallel Lines

Parallel lines are lines that never intersect. They have the same steepness.

Condition: If line 1 has gradient \(m_1\) and line 2 has gradient \(m_2\), then they are parallel if:

\[m_1 = m_2\]

Example: The line \(y = 3x + 1\) is parallel to \(y = 3x - 5\). Both gradients are \(m=3\).

2.2 Perpendicular Lines

Perpendicular lines intersect at a right angle (\(90^\circ\)). Their gradients have a special inverse relationship.

Condition: If line 1 has gradient \(m_1\) and line 2 has gradient \(m_2\), then they are perpendicular if:

\[m_1 m_2 = -1 \quad \text{or} \quad m_2 = - \frac{1}{m_1}\]

This means the perpendicular gradient is the negative reciprocal of the original gradient.

Mnemonic: To find the negative reciprocal: 1. Flip the fraction. 2. Change the sign.

Example: If \(m_1 = \frac{2}{5}\), the perpendicular gradient \(m_2\) is \(-\frac{5}{2}\).

⚠️ Common Error Alert: Do not forget to do both steps (flip and negate)! A common mistake is just taking the negative gradient (\(-m_1\)) instead of the negative reciprocal (\(-\frac{1}{m_1}\)).

3. Coordinate Geometry: Length, Midpoint, and Bisectors

We use straight-line geometry to measure distance and locate points along a line segment defined by two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\).

3.1 Length of a Line Segment (Distance)

This is simply the application of the Pythagoras Theorem on the coordinate grid:

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

3.2 Midpoint of a Line Segment

The midpoint \(M\) is the exact center, found by calculating the average of the x-coordinates and the average of the y-coordinates:

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

3.3 The Perpendicular Bisector (A Comprehensive Problem)

A perpendicular bisector is a line that cuts a segment into two equal halves (uses the midpoint) AND is perpendicular to the segment (uses the negative reciprocal gradient).

Finding the equation of a perpendicular bisector is a common, multi-step exam question. Follow these steps:

1. Find the Midpoint (M): Use the midpoint formula from 3.2. This gives you a point \((x_1, y_1)\) for your new line.

2. Find the Gradient (\(m_{\text{seg}}\)): Calculate the gradient of the original line segment.

3. Find the Perpendicular Gradient (\(m_{\text{perp}}\)): Use the condition \(m_{\text{perp}} = - \frac{1}{m_{\text{seg}}}\).

4. Find the Equation: Use the Point-Gradient Form, \(y - y_1 = m_{\text{perp}}(x - x_1)\), substituting the Midpoint (M) and the perpendicular gradient.

Don't worry if this seems tricky at first—it’s just three formulas used in sequence! Practice makes perfect.

4. Transforming Non-Linear Relationships to Linear Form (Logarithmic and Power Laws)

This is where Additional Mathematics truly uses the straight line concept to solve complex problems that involve constants or exponents. The goal is to take a difficult equation and plot it as a simple straight line \(Y = mX + C\).

4.1 The Core Idea: \(Y = mX + C\)

In this transformation:

1. The original variables \(x\) and \(y\) are replaced by new variables \(X\) and \(Y\). Often \(X\) or \(Y\) involve logarithms, or powers of \(x\) or \(y\).

2. The constants we are trying to find (like \(A\), \(n\), or \(b\)) become the new gradient (\(m\)) or Y-intercept (\(C\)).

4.2 Case 1: Power Law Relationships (Form \(y = Ax^n\))

If you have an equation where one variable is raised to a power, like \(y = Ax^n\), you cannot plot \(y\) against \(x\) as a straight line. We use logarithms (usually \(\log_{10}\) or \(\ln\)) to linearize it.

Transformation Process:

Apply \(\log\) to both sides:

\[\log y = \log (Ax^n)\]

Use the laws of logarithms (\(\log (AB) = \log A + \log B\) and \(\log x^n = n \log x\)):

\[\log y = \log A + n \log x\]

Comparing to \(Y = mX + C\):

    \(\log y\)   =   \(n\) \(\log x\)   +   \(\log A\)
      \(Y\)     =   \(m\)     \(X\)     +     \(C\)

• New Y-Axis (Y): \(\log y\)
• New X-Axis (X): \(\log x\)
• Gradient (m): \(n\) (the power)
• Y-intercept (C): \(\log A\) (where A is the coefficient)

If you plot \(\log y\) against \(\log x\) and get a straight line, you can find the unknown constants \(n\) and \(A\) by finding the gradient and intercept of this new line!

4.3 Case 2: Exponential Relationships (Form \(y = Ab^x\))

If the variable \(x\) is in the exponent, we also use logarithms.

Transformation Process:

Apply \(\log\) to both sides:

\[\log y = \log (Ab^x)\]

Using log laws:

\[\log y = \log A + x \log b\]

Comparing to \(Y = mX + C\):

    \(\log y\)   =   \((\log b)\)   \(x\)   +   \(\log A\)
      \(Y\)     =       \(m\)         \(X\)   +     \(C\)

• New Y-Axis (Y): \(\log y\)
• New X-Axis (X): \(x\) (note that \(X\) is just the original \(x\))
• Gradient (m): \(\log b\)
• Y-intercept (C): \(\log A\)

4.4 Case 3: Relationships Already in Linear Form

Sometimes, the transformation involves algebraic manipulation rather than logarithms, resulting in a form that is already a straight line if plotted correctly.

Example: If the relationship is \(y^2 = Ax^3 + B\).

This is already in the form \(Y = mX + C\):

\[(y^2) = (A) (x^3) + (B)\]

• New Y-Axis (Y): \(y^2\)
• New X-Axis (X): \(x^3\)
• Gradient (m): \(A\)
• Y-intercept (C): \(B\)

Did you know? This transformation technique is incredibly useful in experimental science (like Physics or Chemistry) to check if your data follows a theoretical relationship. Scientists plot the linearized axes to verify their models!