4MA1 Transformation Geometry

Write down the coordinates of the image of the point $(2, -5)$ after a reflection in the line $x = 4$. [1]

Why this scores

Mark scheme:

• B1 for $(6, -5)$

Why this scores full marks: The script recognises $x = 4$ as a vertical line, so only the $x$-coordinate changes. The usual wrong answers are $(2, 13)$, from reflecting in the horizontal line $y = 4$, and $(-2, -5)$, from reflecting in the $y$-axis.

Shape $S$ is translated by the vector $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ to give shape $T$.

Shape $T$ is translated by the vector $\begin{pmatrix} -5 \\ 4 \end{pmatrix}$ to give shape $U$.

Describe fully the single transformation that maps shape $S$ onto shape $U$. [2]

Why this scores

Mark scheme:

• B1 for translation
• B1 for $\begin{pmatrix} -2 \\ 2 \end{pmatrix}$

Why this scores full marks: One translation followed by another is a single translation, and its vector is the sum of the two vectors. The script names the transformation as well as giving the vector; the vector alone scores B1.

Shape $A$ has an angle of 40° and a side of length $5$ cm.

Shape $A$ is enlarged by scale factor $3$ to give shape $B$.

(a) Write down the size of the matching angle in shape $B$. [1]

(b) Write down the length of the matching side in shape $B$. [1]

Why this scores

Mark scheme:

• B1 for 40°
• B1 for $15$ cm

Why this scores full marks: The script separates what an enlargement changes (lengths, by the scale factor) from what it keeps (angles, so $A$ and $B$ are similar). Reflections, rotations and translations keep both lengths and angles, so their images are congruent; an enlargement with a scale factor other than $1$ does not.

Triangle $R$ has vertices $(3, -2)$, $(5, -2)$ and $(5, 1)$.

Triangle $R$ is rotated 90° anticlockwise about the point $(1, -2)$.

Find the coordinates of the vertices of the image. [3]

Why this scores

Mark scheme:

• B3 for all three vertices correct (B2 for two correct, or for a correct rotation of 90° clockwise about $(1, -2)$; B1 for a correct rotation of 90° anticlockwise about any other centre)

Why this scores full marks: The script measures every vector from the centre $(1, -2)$, turns it, and adds it back on to the centre. The side along $y = -2$ points right from the centre, so after a quarter turn anticlockwise it points straight up, a quick check on the first two vertices. Rotating about the origin instead gives $(2, 3)$, $(2, 5)$ and $(-1, 5)$.

The point $A(-1, 4)$ is reflected to the point $A'(5, 4)$.

(a) Find the equation of the mirror line. [1]

(b) Write down the coordinates of the image of the point $(0, -3)$ under the same reflection. [1]

(c) Write down the coordinates of the point on the $x$-axis that is invariant under this reflection. [1]

Why this scores

Mark scheme:

• B1 for $x = 2$
• B1 for $(4, -3)$
• B1 for $(2, 0)$

Why this scores full marks: The script uses the fact that the mirror line is halfway between a point and its image, at right angles to the line joining them. Points on the mirror line are invariant, so the invariant point on the $x$-axis is where $x = 2$ crosses it. Writing "$x = 2$" as "2" or "the line 2" loses the mark in (a).

Triangle $A$ has vertices $(-4, 5)$, $(2, 5)$ and $(2, -1)$.

Triangle $B$ has vertices $(-3, 3)$, $(0, 3)$ and $(0, 0)$.

Describe fully the single transformation that maps triangle $A$ onto triangle $B$. [3]

Why this scores

Mark scheme:

• B3 for enlargement, scale factor $\dfrac{1}{2}$, centre $(-2, 1)$ (B2 for two of these, B1 for one)

Why this scores full marks: The script finds the scale factor by dividing an image length by the matching object length, then finds the centre and checks it with a second pair of vertices. On a grid the same centre comes from drawing lines through matching vertices. Giving scale factor $2$, which maps $B$ onto $A$, is the common error.

(a) Which of the points $(3, 3)$, $(3, -3)$, $(0, 0)$ and $(-2, -2)$ are invariant under a reflection in the line $y = x$? [1]

(b) Write down the coordinates of the invariant point of a rotation of 90° anticlockwise about $(1, 4)$. [1]

(c) Explain why a translation by the vector $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$ has no invariant points. [1]

(d) Write down the coordinates of the invariant point, under a reflection in the line $x = -3$, whose $y$-coordinate is $7$. [1]

Why this scores

Mark scheme:

• B1 for $(3, 3)$, $(0, 0)$ and $(-2, -2)$ and no others
• B1 for $(1, 4)$
• B1 for a correct explanation, e.g. every point is moved by the same non-zero vector
• B1 for $(-3, 7)$

Why this scores full marks: The script knows where each transformation's invariant points are: every point on the mirror line for a reflection, only the centre for a rotation, and none at all for a translation by a non-zero vector. $(3, -3)$ is on $y = -x$, not $y = x$, so it moves to $(-3, 3)$.

Triangle $P$ has vertices $(1, 3)$, $(4, 3)$ and $(4, 5)$.

(a) Triangle $P$ is reflected in the line $y = 2$ to give triangle $Q$. Write down the coordinates of the vertices of triangle $Q$. [1]

(b) Triangle $Q$ is reflected in the line $x = -1$ to give triangle $R$. Write down the coordinates of the vertices of triangle $R$. [1]

(c) Describe fully the single transformation that maps triangle $P$ onto triangle $R$. [3]

Why this scores

Mark scheme:

• B1 for $(1, 1)$, $(4, 1)$, $(4, -1)$
• B1 ft for $(-3, 1)$, $(-6, 1)$, $(-6, -1)$
• B3 for rotation, 180°, centre $(-1, 2)$ (B2 for two of these, B1 for one)

Why this scores full marks: Reflections in two perpendicular lines combine to a half turn about the point where the lines cross, here $(-1, 2)$. The script proves it with midpoints rather than relying on the shape looking upside down. "Reflection in $y = 2$ then $x = -1$" is two transformations and scores 0 in (c).

Triangle $A$ has vertices $(-6, -3)$, $(0, -3)$ and $(0, 6)$.

Triangle $A$ is enlarged by scale factor $\dfrac{1}{3}$, centre $(3, 3)$, to give triangle $B$.

(a) Find the coordinates of the vertices of triangle $B$. [2]

(b) Describe fully the single transformation that maps triangle $B$ onto triangle $A$. [3]

Why this scores

Mark scheme:

• M1 for a correct vector from $(3, 3)$ multiplied by $\dfrac{1}{3}$, or for two correct vertices
• A1 for $(0, 1)$, $(2, 1)$, $(2, 4)$
• B3 for enlargement, scale factor $3$, centre $(3, 3)$ (B2 for two of these, B1 for one)

Why this scores full marks: The inverse of an enlargement keeps the same centre and uses the reciprocal scale factor. The script checks it: from $(3, 3)$ to $(2, 4)$ is $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$, and three times that reaches $(0, 6)$. Giving scale factor $\dfrac{1}{3}$ again in (b) is the common error.

Triangle $A$ has vertices $(1, 1)$, $(4, 1)$ and $(4, 3)$.

Triangle $B$ has vertices $(-2, 0)$, $(-2, -3)$ and $(0, -3)$.

(a) Describe fully the single transformation that maps triangle $A$ onto triangle $B$. [3]

(b) Find the coordinates of the image of the point $(0, 5)$ under the same transformation. [2]

Why this scores

Mark scheme:

• B3 for rotation, 90° clockwise (or 270° anticlockwise), centre $(-1, 2)$ (B2 for two of these, B1 for one)
• M1 for a vector from their centre to $(0, 5)$ turned 90° clockwise, e.g. $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$
• A1 for $(2, 1)$

Why this scores full marks: With no grid, the script finds the centre algebraically: it is equidistant from each point and its image, so it lies on two perpendicular bisectors. A check with the third vertex confirms it, since from $(-1, 2)$ to $(4, 3)$ is $\begin{pmatrix} 5 \\ 1 \end{pmatrix}$, which turns to $\begin{pmatrix} 1 \\ -5 \end{pmatrix}$ and lands on $(0, -3)$.