A local market trader has a triangular stall sign, shape \(K\), with vertices \((1,2)\), \((4,2)\) and \((1,5)\), shown on a plan of the market square. Shap...
Question 1 Report
A local market trader has a triangular stall sign, shape \(K\), with vertices \((1,2)\), \((4,2)\) and \((1,5)\), shown on a plan of the market square. Shape \(K\) is rotated \(90^{\circ}\) clockwise about the point \((1,2)\) to give shape \(K'\).
- Find the coordinates of the image of \((4,2)\). (2)
- Find the coordinates of the image of \((1,5)\). (2)
Answer Details
Rotating a point \((x,y)\) by \(90^\circ\) clockwise about a centre \((a,b)\) is done by finding the point's position relative to the centre, applying the rule \((p,q)\to(q,-p)\) to that offset, then adding the centre back on.
- The point \((4,2)\) relative to the centre \((1,2)\) is \((3,0)\). Rotating \(90^\circ\) clockwise: \((3,0)\to(0,-3)\). Adding the centre back: \((1,2)+(0,-3)=(1,-1)\). So the image of \((4,2)\) is \((1,-1)\). [2 marks]
- The point \((1,5)\) relative to the centre \((1,2)\) is \((0,3)\). Rotating \(90^\circ\) clockwise: \((0,3)\to(3,0)\). Adding the centre back: \((1,2)+(3,0)=(4,2)\). So the image of \((1,5)\) is \((4,2)\). [2 marks]
Exam tip: always subtract the centre before applying the rotation rule and add it back afterwards; applying the rule directly to the original coordinates only works when the centre of rotation is the origin.
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