2.2.3 Invariant Points & Lines

Overview

Invariant Points

A point is invariant under a linear transformation if its coordinates remain unchanged after the transformation is applied. Mathematically, for a matrix $M$ , a point $\mathbf{x} = (x, y)$ is invariant if $M \mathbf{x} = \mathbf{x}$

Invariant Lines

A line is invariant under a linear transformation if any point on the line is transformed to another point on the same line. For a line $y = mx + c$ , it is invariant if:

M \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix},

where $y′=mx′+c$

Finding Invariant Points

Start with the Given Transformation Matrix $M$ :

Write the transformation equations:

\begin{pmatrix} x' \\ y' \end{pmatrix} = M \begin{pmatrix} x \\ y \end{pmatrix}.

Equate this to $\begin{pmatrix} x \\ y \end{pmatrix}$

M \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}.

This gives two equations to solve.

Simplify and Solve the Equations:

Rearrange the equations to isolate $x$ and $y$ , then solve simultaneously.

Interpret the Results:

A single solution corresponds to a specific invariant point.
Infinite solutions describe a line of invariant points.

Finding Invariant Lines

Start with the General Line Equation:

Use:

y = mx + c

Substitute into the Transformation:

Replace $y$ with $mx+c$ in the transformation matrix:

\begin{pmatrix} x' \\ y' \end{pmatrix} = M \begin{pmatrix} x \\ mx + c \end{pmatrix}

Simplify and Equate:

The transformed $y′$ must satisfy $y′=mx′+c$ . Rearrange to form equations for $m$ and $c$ .

Analyze the Conditions:

Solve for $m$ (the slope) and $c$ (the intercept) based on consistent equations.

Worked Examples

infoNote

Example 1: Finding Invariant Points

Given Matrix

Let:

M = \begin{pmatrix} 4.2 & 1.6 \\ 1.6 & 1.8 \end{pmatrix}

Step 1: Write the Transformation Equations

From

M \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}

We get:

\begin{aligned} 4.2x + 1.6y &= x, \\ 1.6x + 1.8y &= y. \end{aligned}

Step 2: Rearrange Equations

Rewriting each equation:

3.2x + 1.6y = 0 \quad \text{(Equation 1)},

1.6x + 0.8y = 0 \quad \text{(Equation 2)}.

Step 3: Solve Simultaneously

Divide Equation $2$ by $0.8$ :

2x + y = 0 \quad \Rightarrow \quad y = -2x.

Step 4: Interpret the Solution

The invariant points form the line:

y = -2x.

infoNote

Example 2: Finding Invariant Lines

Given Matrix

Let:

M = \begin{pmatrix} 7 & 24 \\ 24 & -7 \end{pmatrix}.

Step 1: Start with General Line Equation

Assume:

y = mx + c.

Step 2: Apply the Transformation

Substitute $y = mx + c$ into the transformation:

\begin{pmatrix} x' \\ y' \end{pmatrix} = M \begin{pmatrix} x \\ mx + c \end{pmatrix}.

This expands to:

\begin{aligned} x' &= 7x + 24(mx + c), \\ y' &= 24x - 7(mx + c). \end{aligned}

Step 3: Expand and Equate

Using $y' = mx' + c$ , substitute $x'$ and $y'$ :

24x - 7(mx + c) = m(7x + 24mx + 24c) + c.

Step 4: Simplify

Gather terms involving $x$ and $c$ :

0 = 2x(12m + 7m - 12) + 8c(1 + 3m).

Step 5: Solve for $m$ and $c$

If $c = 0$ , the quadratic simplifies, giving:

m = \frac{4}{3} \quad \text{or} \quad m = -\frac{3}{4}.

Step 6: Write the Invariant Lines

The invariant lines are:

y = \frac{4}{3}x \quad \text{and} \quad y = -\frac{3}{4}x.

Note Summary

infoNote

Common Mistakes

Confusing Invariant Points with Invariant Lines: Invariant points remain unchanged, while invariant lines may map to themselves but change points along the line.
Incorrect Setup: Forgetting to use $M \mathbf{x} = \mathbf{x}$ for points or $y′=mx′+c$ for lines.
Special Cases: Missing c = 0 or symmetrical properties of the transformation matrix.
Verification Errors: Failing to substitute solutions back into the original equations.
Quadratic Errors: Miscalculating the values of $m$ or $c$ .

infoNote

Key Formulas

Invariant Point Condition:

M \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}.

Invariant Line General Form: $y=mx+c$ , where $y′=mx′+c$
Matrix Transformation for Line Verification:

M \begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix}

Slope Conditions: Solve the quadratic equation derived from $M$ :

0 = 2x(12m + 7m - 12) + 8c(1 + 3m).

Intercept Conditions: If c = 0, this simplifies to specific invariant lines:

y = \frac{4}{3}x\ \text{or}\ y = -\frac{3}{4}x.

Invariant Points & Lines Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

2.2.3 Invariant Points & Lines

Overview

Invariant Points

Invariant Lines

Finding Invariant Points

Start with the Given Transformation Matrix $M$ :

Simplify and Solve the Equations:

Interpret the Results:

Finding Invariant Lines

Start with the General Line Equation:

Substitute into the Transformation:

Simplify and Equate:

Analyze the Conditions:

Worked Examples

Example 1: Finding Invariant Points

Example 2: Finding Invariant Lines

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Invariant Points & Lines For their A-Level Exams.

Other Revision Notes related to Invariant Points & Lines you should explore

Invariant Points & Lines Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

2.2.3 Invariant Points & Lines

Overview

Invariant Points

Invariant Lines

Finding Invariant Points

Start with the Given Transformation Matrix MMM:

Simplify and Solve the Equations:

Interpret the Results:

Finding Invariant Lines

Start with the General Line Equation:

Substitute into the Transformation:

Simplify and Equate:

Analyze the Conditions:

Worked Examples

Example 1: Finding Invariant Points

Example 2: Finding Invariant Lines

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Invariant Points & Lines For their A-Level Exams.

Other Revision Notes related to Invariant Points & Lines you should explore

Start with the Given Transformation Matrix $M$ :