Reading the vertices off the grid first makes every part straightforward. From the diagram, \(P\) has vertices \((1, 1)\), \((3, 1)\) and \((1, 4)\); \(Q\) has \((5, 1)\), \((7, 1)\) and \((5, 4)\); and \(R\) has \((-1, 1)\), \((-3, 1)\) and \((-1, 4)\).
(a) Comparing corresponding vertices, \((1, 1) \to (5, 1)\), \((3, 1) \to (7, 1)\) and \((1, 4) \to (5, 4)\). Every point moves \(4\) to the right and \(0\) up, and the triangle keeps the same orientation and size. That is a translation. [1]
The translation vector is \(\begin{pmatrix} 4 \\ 0 \end{pmatrix}\). [1]
A translation is described fully by its vector alone; no centre or line is needed.
(b) Here \((1, 1) \to (-1, 1)\), \((3, 1) \to (-3, 1)\) and \((1, 4) \to (-1, 4)\). The \(y\)-coordinates are unchanged while the \(x\)-coordinates change sign, and the triangle is turned over rather than slid. That is a reflection. [1]
The mirror line is the \(y\)-axis, the line \(x = 0\). [1]
A reflection is described fully by naming the mirror line, so both the word and the line are needed for the two marks.
(c) Reflecting in the \(x\)-axis leaves \(x\) unchanged and reverses the sign of \(y\), so \((x, y) \to (x, -y)\). Applying this to \(R\):
\[(-1, 1) \to (-1, -1), \quad (-3, 1) \to (-3, -1), \quad (-1, 4) \to (-1, -4)\]
Triangle \(T\) has vertices \((-1, -1)\), \((-3, -1)\) and \((-1, -4)\). [1]
(d) She is correct. A rotation of \(180\)° about the origin sends \((x, y)\) to \((-x, -y)\). Applying that to \(P\):
\[(1, 1) \to (-1, -1), \quad (3, 1) \to (-3, -1), \quad (1, 4) \to (-1, -4)\]
These are exactly the vertices of \(T\) found in part (c), so the rotation does map \(P\) onto \(T\). [1]
This is worth noticing as a general result: a reflection in the \(y\)-axis followed by a reflection in the \(x\)-axis is equivalent to a single \(180\)° rotation about the origin, which is precisely the route \(P \to R \to T\) taken here. A rotation of \(180\)° needs no direction stated, since clockwise and anticlockwise give the same image.