A market trader designs a triangular canopy, shape \(A\), with vertices \((1, 1)\), \((4, 1)\) and \((1, 3)\), to be printed on a stall banner drawn on a co...
Question 1 Report
A market trader designs a triangular canopy, shape \(A\), with vertices \((1, 1)\), \((4, 1)\) and \((1, 3)\), to be printed on a stall banner drawn on a coordinate grid.
- Shape \(A\) is reflected in the line \(y = x\) to give shape \(B\). Find the coordinates of the vertices of \(B\). (2)
- Shape \(B\) is then reflected in the \(x\)-axis to give shape \(C\). Find the coordinates of the vertices of \(C\). (2)
- Describe fully the single transformation that maps shape \(A\) directly onto shape \(C\), and use your vertices from parts (a) and (b) to justify your answer. (2)
Answer Details
Combining two reflections in intersecting lines produces a single rotation about the point where the lines cross, through twice the angle between the lines; here \(y=x\) and the \(x\)-axis meet at the origin at \(45^\circ\), so the combined effect should be a rotation of \(90^\circ\) about \((0,0)\).
- Reflecting in \(y=x\) swaps each pair of coordinates: \((1,1)\to(1,1)\), \((4,1)\to(1,4)\), \((1,3)\to(3,1)\). So \(B\) has vertices \((1,1)\), \((1,4)\) and \((3,1)\). [2 marks]
- Reflecting \(B\) in the \(x\)-axis negates each \(y\)-coordinate: \((1,1)\to(1,-1)\), \((1,4)\to(1,-4)\), \((3,1)\to(3,-1)\). So \(C\) has vertices \((1,-1)\), \((1,-4)\) and \((3,-1)\). [2 marks]
- The single transformation is a rotation of \(90^\circ\) clockwise about \((0,0)\). [1 mark] Checking: this rotation sends \((x,y)\to(y,-x)\), so \(A\)'s vertices \((1,1)\), \((4,1)\), \((1,3)\) map to \((1,-1)\), \((1,-4)\), \((3,-1)\), which are exactly the vertices of \(C\) found in part (b), confirming the two reflections are equivalent to this single rotation. [1 mark]
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