AS & A Level · Practice questions · Invariant points and lines

Find what stays.

Six original Cambridge-style questions on invariant points and invariant lines: solving \(\mathbf{M}\mathbf{x} = \mathbf{x}\), quadratics in \(m\), lines not through the origin and the geometry of a shear.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Map a general point, then ask where its image lies.

Invariant points: solve \(\mathbf{M}\mathbf{x} = \mathbf{x}\), two equations through \(O\); a line of them exists exactly when \(\det(\mathbf{M} - \mathbf{I}) = 0\).
Lines through \(O\): map \((x, mx)\) to \((X, Y)\) and require \(Y = mX\); solve the quadratic in \(m\).
Check \(x = 0\) separately; it is invariant exactly when the top-right entry is 0.
Lines \(y = mx + c\): require \(Y = mX + c\) for all \(x\) and compare BOTH the \(x\)-coefficients and the constants.
Classify each invariant line: test one point to see whether it stays put or slides along the line.

Recall

[2 marks]

Find the invariant points of the transformation with matrix \(\begin{pmatrix} 4 & -3 \\ 2 & -1 \end{pmatrix}\).

\(4x - 3y = x\) and \(2x - y = y\) give \(3x - 3y = 0\) and \(2x - 2y = 0\). ✓
Both are \(y = x\): every point on the line \(y = x\) is invariant (check \((1, 1) \mapsto (1, 1)\)). ✓

Calculation

[4 marks]

\(\mathbf{M} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\).

(a) Show that the origin is the only invariant point of \(\mathbf{M}\).

(b) Find the invariant lines through the origin, giving the gradients in exact form.

(a) \(2x + 5y = x\) and \(x + 3y = y\) give \(x + 5y = 0\) and \(x + 2y = 0\). ✓
Subtracting, \(3y = 0\), so \(y = 0\) and \(x = 0\). (Equivalently \(\det(\mathbf{M} - \mathbf{I}) = 1 \times 2 - 5 \times 1 = -3 \ne 0\).) ✓
(b) \((x, mx) \mapsto \big((2 + 5m)x, (1 + 3m)x\big)\); \(1 + 3m = m(2 + 5m)\) gives \(5m^{2} - m - 1 = 0\). ✓
\(m = \dfrac{1 \pm \sqrt{1 + 20}}{10} = \dfrac{1 \pm \sqrt{21}}{10}\), so the lines are \(y = \dfrac{1 + \sqrt{21}}{10}x\) and \(y = \dfrac{1 - \sqrt{21}}{10}x\) (and \(x = 0\) is not invariant, since the top-right entry is \(5 \ne 0\)). ✓

Calculation

[4 marks]

Find the invariant lines through the origin for the transformation with matrix \(\begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix}\). Show that neither line is a line of invariant points.

\((x, mx) \mapsto \big((1 + 4m)x, (2 + 3m)x\big)\); \(2 + 3m = m(1 + 4m)\) gives \(4m^{2} - 2m - 2 = 0\). ✓
\(2(2m + 1)(m - 1) = 0\), so \(m = 1\) or \(m = -\tfrac{1}{2}\): the lines \(y = x\) and \(y = -\tfrac{1}{2}x\). ✓
\((1, 1) \mapsto (5, 5)\), which is on \(y = x\) but is not \((1, 1)\). ✓
\((2, -1) \mapsto (-2, 1)\), which is on \(y = -\tfrac{1}{2}x\) but is not \((2, -1)\). So both lines are invariant, but their points move. ✓

Analysis

[4 marks]

\(\mathbf{M} = \begin{pmatrix} k & 1 \\ 6 & 1 \end{pmatrix}\). The line \(y = 2x\) is invariant under the transformation represented by \(\mathbf{M}\).

(a) Find \(k\).

(b) Find the other invariant line through the origin.

(a) \((x, 2x) \mapsto \big((k + 2)x, 8x\big)\). For the image to be on \(y = 2x\): \(8x = 2(k + 2)x\), so \(k + 2 = 4\) and \(k = 2\). ✓
(b) With \(\mathbf{M} = \begin{pmatrix} 2 & 1 \\ 6 & 1 \end{pmatrix}\): \((x, mx) \mapsto \big((2 + m)x, (6 + m)x\big)\). ✓
\(6 + m = m(2 + m)\) gives \(m^{2} + m - 6 = 0\), \((m + 3)(m - 2) = 0\). ✓
\(m = 2\) is the given line, so the other invariant line is \(y = -3x\) (check \((1, -3) \mapsto (-1, 3)\)). ✓

Analysis

[5 marks]

\(\mathbf{M} = \begin{pmatrix} -1 & 1 \\ -4 & 3 \end{pmatrix}\).

(a) Show that every line with gradient 2 is an invariant line of the transformation represented by \(\mathbf{M}\).

(b) Find the invariant points of the transformation.

(a) The point \((x, 2x + c)\) maps to \(X = -x + (2x + c) = x + c\) and \(Y = -4x + 3(2x + c) = 2x + 3c\). ✓
\(2X + c = 2x + 2c + c = 2x + 3c = Y\), so the image lies on \(y = 2x + c\) for every \(c\). ✓
(b) \(-x + y = x\) and \(-4x + 3y = y\) give \(y = 2x\) and \(4y = 8x\), both \(y = 2x\): every point of \(y = 2x\) is invariant. ✓
(c) On \(y = 2x\) each point stays where it is. \(y = 2x + 1\) is invariant as a line, but its points move along it: \((0, 1) \mapsto (1, 3)\), which is on \(y = 2x + 1\) but is a different point. ✓

Proof

[5 marks]

The matrix \(\mathbf{S} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\), where \(k \ne 0\), represents a shear.

(a) Show that the invariant points of \(\mathbf{S}\) are exactly the points of the \(x\)-axis.

(b) Prove that the invariant lines of \(\mathbf{S}\) are exactly the lines \(y = c\).

(a) \(x + ky = x\) and \(y = y\). The first gives \(ky = 0\), and \(k \ne 0\), so \(y = 0\). Every point \((x, 0)\) is invariant, and no other point is. ✓
(b) Vertical lines \(x = h\): \((h, y) \mapsto (h + ky, y)\), which leaves \(x = h\) whenever \(y \ne 0\), so no vertical line is invariant. ✓
Other lines \(y = mx + c\): \((x, mx + c) \mapsto \big(x + k(mx + c),\ mx + c\big)\). Require \(Y = mX + c\): \(mx + c = m(1 + km)x + kmc + c\) for all \(x\). ✓
Coefficients of \(x\): \(m = m + km^{2}\), so \(km^{2} = 0\) and \(m = 0\). Constants: \(c = kmc + c\), true when \(m = 0\) for every \(c\). ✓
So the invariant lines are exactly \(y = c\): each point \((x, c)\) slides to \((x + kc, c)\) along its own horizontal line. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.