IGCSE Mathematics: Vectors and Transformations

Cambridge IGCSE 0580 & Edexcel 4MA1 · Updated March 2026

This chapter covers all aspects of geometric transformations — translations, reflections, rotations, and enlargements — as well as an introduction to vector notation and vector geometry. Transformations describe how shapes move or change in the coordinate plane, and vectors provide the algebraic language to describe direction and magnitude precisely.

Specification Note

Content labelled Extended is required for Extended tier (Cambridge) or Higher tier (Edexcel) only.

1. Translations

A translation moves every point of a shape by the same distance in the same direction. The shape does not rotate, reflect, or change size — it simply slides to a new position. Translations are described using a column vector.

Column Vector

A column vector $\begin{pmatrix} a \\ b \end{pmatrix}$ means: move $a$ units in the $x$-direction (positive = right, negative = left) and $b$ units in the $y$-direction (positive = up, negative = down).

Properties of Translations

The object and its image are congruent (same shape and size).
Corresponding sides of the object and image are parallel.
No point is fixed (no invariant points, unless the vector is $\begin{pmatrix}0\\0\end{pmatrix}$).
The translation that maps the image back to the object is $\begin{pmatrix}-a\\-b\end{pmatrix}$.

Example 1 — Translating a triangle

Triangle $ABC$ has vertices $A(1, 2)$, $B(3, 2)$, and $C(2, 4)$. Translate the triangle by $\begin{pmatrix}3 \\ -1\end{pmatrix}$.

Step 1 Add the column vector to each vertex:

$A' = (1+3,\ 2+(-1)) = (4, 1)$

$B' = (3+3,\ 2+(-1)) = (6, 1)$

$C' = (2+3,\ 4+(-1)) = (5, 3)$

Step 2 Plot $A'(4,1)$, $B'(6,1)$, $C'(5,3)$ and join them to form triangle $A'B'C'$.

The image $A'B'C'$ is congruent to $ABC$, moved 3 right and 1 down.

Example 2 — Describing a translation

Shape $P$ has a vertex at $(5, 6)$. The image $P'$ has the corresponding vertex at $(2, 9)$. Describe the translation.

Step 1 Find the change in each coordinate:

$x$-change: $2 - 5 = -3$

$y$-change: $9 - 6 = +3$

Step 2 Write as a column vector: Translation is $\begin{pmatrix}-3 \\ 3\end{pmatrix}$.

Practice 1a

Quadrilateral $PQRS$ has vertices $P(0,0)$, $Q(2,0)$, $R(2,3)$, $S(0,3)$. Find the image vertices after a translation by $\begin{pmatrix}-1 \\ 4\end{pmatrix}$.

Show Solution

Add $\begin{pmatrix}-1\\4\end{pmatrix}$ to each vertex:

$P' = (0-1,\ 0+4) = (-1, 4)$

$Q' = (2-1,\ 0+4) = (1, 4)$

$R' = (2-1,\ 3+4) = (1, 7)$

$S' = (0-1,\ 3+4) = (-1, 7)$

Practice 1b

A triangle has a vertex at $(4, -2)$. After a translation, that vertex is at $(-1, 5)$. Write the translation vector.

Show Solution

$x$-change: $-1 - 4 = -5$

$y$-change: $5 - (-2) = 7$

Translation: $\begin{pmatrix}-5 \\ 7\end{pmatrix}$

2. Reflections

A reflection creates a mirror image of a shape across a mirror line. Each point and its image are equidistant from the mirror line, and the line joining them is perpendicular to the mirror line.

Properties of Reflections

Object and image are congruent.
Points on the mirror line are invariant (they do not move).
The object and image are on opposite sides of the mirror line.
Orientation is reversed (left and right are swapped).

Key Mirror Lines

$x$-axis ($y = 0$): $(x, y) \to (x, -y)$
$y$-axis ($x = 0$): $(x, y) \to (-x, y)$
$y = x$: $(x, y) \to (y, x)$
$y = -x$: $(x, y) \to (-y, -x)$
$x = a$ (vertical line): $(x, y) \to (2a - x, y)$
$y = b$ (horizontal line): $(x, y) \to (x, 2b - y)$

Figure 7.1 — Triangle $ABC$ (blue) and its reflection in the $y$-axis (red). Each point is the same distance from the mirror line on the opposite side.

Example 3 — Reflecting in $y = x$

Reflect triangle with vertices $A(1, 3)$, $B(4, 3)$, $C(4, 6)$ in the line $y = x$.

Step 1 Apply the rule $(x, y) \to (y, x)$:

$A(1, 3) \to A'(3, 1)$

$B(4, 3) \to B'(3, 4)$

$C(4, 6) \to C'(6, 4)$

Step 2 Plot and join $A'$, $B'$, $C'$. Verify by checking each image point is the same distance from $y = x$ as the original, on the opposite side.

Example 4 — Describing a reflection

Shape $P$ has a vertex at $(3, 5)$ and its image $P'$ has the corresponding vertex at $(3, -1)$. Identify the mirror line.

Step 1 The $x$-coordinate is unchanged, so the mirror line is horizontal.

Step 2 The midpoint of $(3, 5)$ and $(3, -1)$ gives the mirror line's $y$-value:

Midpoint $y$-value $= \dfrac{5 + (-1)}{2} = \dfrac{4}{2} = 2$

Mirror line: $y = 2$

Practice 2a

Reflect the point $(-3, 4)$ in (i) the $x$-axis, (ii) the line $y = -x$.

Show Solution

(i) Reflection in $x$-axis: $(x,y) \to (x,-y)$, so $(-3, 4) \to (-3, -4)$.

(ii) Reflection in $y = -x$: $(x,y) \to (-y,-x)$, so $(-3, 4) \to (-4, 3)$.

Practice 2b

A vertex of shape $Q$ is at $(6, 2)$. After reflection, its image is at $(-2, 2)$. Write the equation of the mirror line.

Show Solution

The $y$-coordinate is unchanged, so the mirror line is vertical.

Midpoint $x$-value $= \dfrac{6 + (-2)}{2} = \dfrac{4}{2} = 2$

Mirror line: $x = 2$

3. Rotations

A rotation turns a shape through an angle about a fixed point called the centre of rotation. Angles are measured anticlockwise unless stated otherwise. To describe a rotation fully, state: the centre, the angle, and the direction.

Properties of Rotations

Object and image are congruent.
The centre of rotation is the only invariant point.
Every point on the object moves through the same angle.
Orientation is preserved (same sense as a reflection reversal does not apply).

Common Rotations About the Origin

90° anticlockwise: $(x, y) \to (-y, x)$
90° clockwise: $(x, y) \to (y, -x)$
180° (either direction): $(x, y) \to (-x, -y)$

Example 5 — Rotating 90° anticlockwise about the origin

Triangle $DEF$ has vertices $D(2, 1)$, $E(5, 1)$, $F(5, 4)$. Rotate it 90° anticlockwise about the origin.

Step 1 Apply $(x, y) \to (-y, x)$:

$D(2, 1) \to D'(-1, 2)$

$E(5, 1) \to E'(-1, 5)$

$F(5, 4) \to F'(-4, 5)$

Step 2 Plot and join $D'$, $E'$, $F'$. Check that each point is the same distance from the origin as the corresponding original point.

Example 6 — Finding the centre of rotation

Triangle $P$ with vertex at $(4, 0)$ is mapped to $P'$ with vertex at $(0, 4)$. A second vertex of $P$ is at $(4, 2)$, mapped to $(−2, 4)$ in $P'$. Find the centre and angle of rotation.

Step 1 Draw the perpendicular bisector of the segment from $(4,0)$ to $(0,4)$. Midpoint = $(2,2)$; gradient of segment = $\dfrac{4-0}{0-4}=-1$; perpendicular gradient = $1$. Bisector: $y - 2 = 1(x - 2) \Rightarrow y = x$.

Step 2 Draw the perpendicular bisector of the segment from $(4,2)$ to $(-2,4)$. Midpoint = $(1,3)$; gradient = $\dfrac{4-2}{-2-4}=-\tfrac{1}{3}$; perpendicular gradient = $3$. Bisector: $y - 3 = 3(x-1) \Rightarrow y = 3x$.

Step 3 Solve $y = x$ and $y = 3x$: $x = 3x \Rightarrow x = 0$, $y = 0$. Centre = $(0,0)$.

Step 4 Check the angle: $(4,0) \to (0,4)$ is a 90° anticlockwise rotation about the origin. ✓

Answer: Rotation of 90° anticlockwise about the origin.

Practice 3a

Point $G(3, 5)$ is rotated 180° about the origin. Write the coordinates of $G'$.

Show Solution

$(x,y) \to (-x,-y)$, so $G(3,5) \to G'(-3,-5)$.

Practice 3b

Rotate the triangle with vertices $(1, 0)$, $(3, 0)$, $(3, 2)$ by 90° clockwise about the origin. State the image vertices.

Show Solution

Apply $(x,y) \to (y,-x)$:

$(1,0) \to (0,-1)$

$(3,0) \to (0,-3)$

$(3,2) \to (2,-3)$

4. Enlargements

An enlargement changes the size of a shape. All lengths are multiplied by the scale factor $k$, and all angles remain the same. The fixed point from which the shape is enlarged is the centre of enlargement.

Properties of Enlargements

Object and image are similar (same shape, different size unless $k = 1$).
All lengths are multiplied by $|k|$.
Area is multiplied by $k^2$.
If $k > 1$: image is larger than object, same orientation.
If $0 < k < 1$: image is smaller than object (reduction), same orientation.
If $k < 0$: image is on the opposite side of the centre, also rotated 180°.

Finding Image Coordinates from a Centre of Enlargement

For centre $C = (c_x, c_y)$ and scale factor $k$:

$$\text{Image point} = C + k \times (\text{object point} - C)$$

Example 7 — Enlargement with positive scale factor

Enlarge triangle $A(1,1)$, $B(3,1)$, $C(3,4)$ by scale factor $2$ from centre $(1,1)$.

Step 1 The centre is $A(1,1)$. Since $A$ is the centre, $A$ maps to itself: $A' = (1,1)$.

Step 2 For $B(3,1)$: vector from centre = $(3-1, 1-1) = (2,0)$. Multiply by $2$: $(4,0)$. Add to centre: $(1+4, 1+0) = (5, 1)$. So $B' = (5,1)$.

Step 3 For $C(3,4)$: vector from centre = $(2,3)$. Multiply by $2$: $(4,6)$. Add to centre: $(5, 7)$. So $C' = (5,7)$.

Verify: $AB = 2$, $A'B' = 4$ ✓; $BC = 3$, $B'C' = 6$ ✓.

Example 8 — Fractional and negative scale factors

(a) Enlarge square with vertices $(2,2)$, $(6,2)$, $(6,6)$, $(2,6)$ by scale factor $\tfrac{1}{2}$ from centre $(0,0)$.

Step 1 Multiply each coordinate by $\tfrac{1}{2}$:

$(2,2) \to (1,1)$; $(6,2) \to (3,1)$; $(6,6) \to (3,3)$; $(2,6) \to (1,3)$

The image is a smaller square of side $2$ (half of $4$).

(b) Enlarge point $P(4, 2)$ by scale factor $-2$ from centre $C(1,1)$.

Step 1 Vector from $C$ to $P$: $(4-1, 2-1) = (3, 1)$.

Step 2 Multiply by $-2$: $(-6, -2)$.

Step 3 Add to $C$: $(1-6, 1-2) = (-5, -1)$. So $P' = (-5, -1)$.

Example 9 — Describing an enlargement

Rectangle $R$ has vertices $(4, 2)$, $(10, 2)$, $(10, 5)$, $(4, 5)$. Its image $R'$ has vertices $(2, 1)$, $(5, 1)$, $(5, 2.5)$, $(2, 2.5)$. Describe the enlargement fully.

Step 1 Scale factor: $\dfrac{\text{image length}}{\text{object length}} = \dfrac{3}{6} = \dfrac{1}{2}$

Step 2 Find the centre by drawing lines through corresponding vertices and finding their intersection. Line through $(4,2)$ and $(2,1)$: gradient $= \dfrac{1-2}{2-4} = \tfrac{1}{2}$; equation $y-2 = \tfrac{1}{2}(x-4) \Rightarrow y = \tfrac{1}{2}x$. This passes through the origin $(0,0)$.

Answer: Enlargement, scale factor $\tfrac{1}{2}$, centre $(0,0)$.

Practice 4a

Triangle $LMN$ has vertices $L(0,0)$, $M(4,0)$, $N(0,3)$. Enlarge it by scale factor $3$ from centre $(0,0)$.

Show Solution

Multiply each coordinate by $3$:

$L' = (0,0)$; $M' = (12,0)$; $N' = (0,9)$

Practice 4b

Shape $S$ has a vertex at $(6, 4)$. After enlargement with centre $(2, 2)$ and scale factor $-1$, find the image of this vertex.

Show Solution

Vector from centre to point: $(6-2, 4-2) = (4, 2)$.

Multiply by $-1$: $(-4, -2)$.

Add to centre: $(2-4, 2-2) = (-2, 0)$.

Image vertex: $(-2, 0)$

5. Combined Transformations

When two transformations are applied in succession, the result may be equivalent to a single transformation. To describe the combined effect, apply the first transformation, then apply the second to the result.

Exam Tip

Always apply transformations in the order stated. "Transformation $A$ followed by transformation $B$" means apply $A$ first, then $B$ to the resulting image.

Example 10 — Two reflections

Triangle $T$ has vertices $(1,1)$, $(3,1)$, $(3,4)$. Apply reflection in the $y$-axis followed by reflection in the $x$-axis. Describe the single equivalent transformation.

Step 1 Reflect in the $y$-axis: $(x,y) \to (-x,y)$

$(1,1) \to (-1,1)$; $(3,1) \to (-3,1)$; $(3,4) \to (-3,4)$

Step 2 Reflect that image in the $x$-axis: $(x,y) \to (x,-y)$

$(-1,1) \to (-1,-1)$; $(-3,1) \to (-3,-1)$; $(-3,4) \to (-3,-4)$

Step 3 Compare with original: $(1,1) \to (-1,-1)$, which is $(x,y) \to (-x,-y)$.

Single equivalent transformation: Rotation of 180° about the origin.

Example 11 — Rotation followed by translation

Point $P(3, 1)$. Apply rotation of 90° anticlockwise about the origin, then translation by $\begin{pmatrix}2\\-3\end{pmatrix}$.

Step 1 Rotate: $(3,1) \to (-1, 3)$

Step 2 Translate: $(-1+2, 3-3) = (1, 0)$

Final image of $P$ is $(1, 0)$.

Practice 5a

Triangle $W$ has vertices $(1,0)$, $(3,0)$, $(3,2)$. Apply reflection in $y = x$, then rotation of 90° clockwise about the origin. Write the final image vertices and identify the single equivalent transformation.

Show Solution

Step 1 — Reflect in $y = x$: $(x,y) \to (y,x)$

$(1,0) \to (0,1)$; $(3,0) \to (0,3)$; $(3,2) \to (2,3)$

Step 2 — Rotate 90° clockwise: $(x,y) \to (y,-x)$

$(0,1) \to (1,0)$; $(0,3) \to (3,0)$; $(2,3) \to (3,-2)$

Compare original to final image:

$(1,0) \to (1,0)$; $(3,0) \to (3,0)$; $(3,2) \to (3,-2)$

This maps $(x,y) \to (x,-y)$: Reflection in the $x$-axis.

6. Vectors Extended

A vector has both magnitude (size) and direction. Vectors are used to describe displacements, velocities, and forces. Unlike a scalar (which has magnitude only), a vector also tells you which way.

Vector Notation

Column vector: $\mathbf{a} = \begin{pmatrix}3\\-2\end{pmatrix}$ — move $3$ right and $2$ down.
Bold notation: $\mathbf{a}$, $\mathbf{b}$, $\mathbf{p}$ (used in print).
Underline notation: $\underline{a}$, $\underline{b}$ (used in handwriting).
Arrow notation: $\overrightarrow{AB}$ — vector from $A$ to $B$.

Magnitude of a Vector

The magnitude (or modulus) of vector $\mathbf{a} = \begin{pmatrix}x\\y\end{pmatrix}$ is:

$$|\mathbf{a}| = \sqrt{x^2 + y^2}$$

Adding and Subtracting Vectors

Add or subtract vectors component by component:

$$\begin{pmatrix}a\\b\end{pmatrix} + \begin{pmatrix}c\\d\end{pmatrix} = \begin{pmatrix}a+c\\b+d\end{pmatrix}$$ $$\begin{pmatrix}a\\b\end{pmatrix} - \begin{pmatrix}c\\d\end{pmatrix} = \begin{pmatrix}a-c\\b-d\end{pmatrix}$$

Multiplying a Vector by a Scalar

$$k \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}ka\\kb\end{pmatrix}$$

Multiplying by a scalar changes the magnitude but not the direction (if $k > 0$). If $k < 0$, the direction is reversed.

Figure 7.2 — Vector addition: $\mathbf{a} + \mathbf{b} = \mathbf{c}$. The vector $\mathbf{c}$ (shown in green) is the diagonal of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$.

Example 12 — Vector arithmetic

Given $\mathbf{a} = \begin{pmatrix}3\\-1\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}-2\\4\end{pmatrix}$, find:

(a) $\mathbf{a} + \mathbf{b}$, (b) $\mathbf{a} - \mathbf{b}$, (c) $3\mathbf{a} - 2\mathbf{b}$, (d) $|\mathbf{a}|$

(a) $\mathbf{a} + \mathbf{b} = \begin{pmatrix}3+(-2)\\-1+4\end{pmatrix} = \begin{pmatrix}1\\3\end{pmatrix}$

(b) $\mathbf{a} - \mathbf{b} = \begin{pmatrix}3-(-2)\\-1-4\end{pmatrix} = \begin{pmatrix}5\\-5\end{pmatrix}$

(c) $3\mathbf{a} - 2\mathbf{b} = 3\begin{pmatrix}3\\-1\end{pmatrix} - 2\begin{pmatrix}-2\\4\end{pmatrix} = \begin{pmatrix}9\\-3\end{pmatrix} - \begin{pmatrix}-4\\8\end{pmatrix} = \begin{pmatrix}13\\-11\end{pmatrix}$

(d) $|\mathbf{a}| = \sqrt{3^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$

Position Vectors

The position vector of point $A$ is $\overrightarrow{OA}$, where $O$ is the origin. If $A = (a_1, a_2)$, then $\overrightarrow{OA} = \begin{pmatrix}a_1\\a_2\end{pmatrix}$.

The vector from $A$ to $B$ is:

$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$$

Vector Geometry

Parallel Vectors

Two vectors are parallel if one is a scalar multiple of the other: $\mathbf{u} \parallel \mathbf{v}$ if $\mathbf{u} = k\mathbf{v}$ for some scalar $k$.

Collinear Points

Points $A$, $B$, $C$ are collinear if $\overrightarrow{AB} = k \cdot \overrightarrow{AC}$ for some scalar $k$. This means $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel and share the point $A$.

Midpoint using Position Vectors

The midpoint $M$ of $AB$ has position vector: $\overrightarrow{OM} = \dfrac{\mathbf{a} + \mathbf{b}}{2}$

Example 13 — Vector geometry proof

$OABC$ is a parallelogram. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $M$ is the midpoint of $AC$ and $N$ is the midpoint of $OB$. Prove that $M = N$ (i.e., the diagonals bisect each other).

Step 1 Since $OABC$ is a parallelogram, $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$.

Step 2 Position vector of $A$: $\mathbf{a}$. Position vector of $C$: $\mathbf{c}$. Midpoint $M$ of $AC$:

$$\overrightarrow{OM} = \frac{\mathbf{a} + \mathbf{c}}{2}$$

Step 3 $N$ is the midpoint of $OB$. $\overrightarrow{OB} = \mathbf{a}+\mathbf{c}$, so:

$$\overrightarrow{ON} = \frac{(\mathbf{a}+\mathbf{c})}{2}$$

Step 4 $\overrightarrow{OM} = \overrightarrow{ON} = \dfrac{\mathbf{a}+\mathbf{c}}{2}$, so $M$ and $N$ are the same point. The diagonals bisect each other. $\square$

Example 14 — Proving collinearity

$O$, $A$, $B$ are points with position vectors $\mathbf{0}$, $\mathbf{a}$, $2\mathbf{a}$ respectively. Show that $O$, $A$, $B$ are collinear.

Step 1 $\overrightarrow{OA} = \mathbf{a}$

Step 2 $\overrightarrow{OB} = 2\mathbf{a} = 2 \cdot \overrightarrow{OA}$

Step 3 Since $\overrightarrow{OB} = 2\overrightarrow{OA}$, the vectors are parallel and share point $O$, so $O$, $A$, $B$ lie on the same straight line. They are collinear. $\square$

Example 15 — Ratio along a line segment

$A$ has position vector $\mathbf{a}$ and $B$ has position vector $\mathbf{b}$. Point $P$ divides $AB$ in the ratio $2:1$. Find the position vector of $P$.

Step 1 $\overrightarrow{AP} = \dfrac{2}{3}\overrightarrow{AB} = \dfrac{2}{3}(\mathbf{b} - \mathbf{a})$

Step 2 Position vector of $P$: $\overrightarrow{OP} = \mathbf{a} + \dfrac{2}{3}(\mathbf{b}-\mathbf{a}) = \mathbf{a} + \dfrac{2\mathbf{b}}{3} - \dfrac{2\mathbf{a}}{3} = \dfrac{\mathbf{a}}{3} + \dfrac{2\mathbf{b}}{3}$

$$\overrightarrow{OP} = \frac{\mathbf{a} + 2\mathbf{b}}{3}$$

Practice 6a

Given $\mathbf{p} = \begin{pmatrix}5\\2\end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix}-1\\3\end{pmatrix}$, find: (i) $\mathbf{p} + \mathbf{q}$, (ii) $2\mathbf{p} - \mathbf{q}$, (iii) $|\mathbf{q}|$.

Show Solution

(i) $\mathbf{p} + \mathbf{q} = \begin{pmatrix}5+(-1)\\2+3\end{pmatrix} = \begin{pmatrix}4\\5\end{pmatrix}$

(ii) $2\mathbf{p} - \mathbf{q} = \begin{pmatrix}10\\4\end{pmatrix} - \begin{pmatrix}-1\\3\end{pmatrix} = \begin{pmatrix}11\\1\end{pmatrix}$

(iii) $|\mathbf{q}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1+9} = \sqrt{10}$

Practice 6b

$O$ is the origin. $A$ has position vector $\begin{pmatrix}4\\2\end{pmatrix}$ and $B$ has position vector $\begin{pmatrix}10\\5\end{pmatrix}$. Show that $O$, $A$, $B$ are collinear.

Show Solution

$\overrightarrow{OA} = \begin{pmatrix}4\\2\end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix}10\\5\end{pmatrix}$.

Check: $\begin{pmatrix}10\\5\end{pmatrix} = \dfrac{5}{2}\begin{pmatrix}4\\2\end{pmatrix}$. ✓

Since $\overrightarrow{OB} = \dfrac{5}{2}\overrightarrow{OA}$, the vectors are parallel and share point $O$.

Therefore $O$, $A$, $B$ are collinear.

Practice 6c

$OABC$ is a quadrilateral where $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $M$ is the midpoint of $OB$ and $N$ is the midpoint of $AC$. Given that $\overrightarrow{AB} = \mathbf{c}$, find the position vector of $M$ and $N$ in terms of $\mathbf{a}$ and $\mathbf{c}$.

Show Solution

$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \mathbf{a} + \mathbf{c}$

Position vector of $M$ (midpoint of $OB$): $\dfrac{\mathbf{a} + \mathbf{c}}{2}$

Position vector of $C$: $\mathbf{c}$. Midpoint $N$ of $AC$:

$\overrightarrow{ON} = \dfrac{\mathbf{a} + \mathbf{c}}{2}$

Since $M$ and $N$ have the same position vector, $M = N$. The diagonals $OB$ and $AC$ bisect each other, so $OABC$ is a parallelogram.