Symmetry Plan for a Community Garden
A community garden committee is using a coordinate grid measured in metres. The existing triangular herb bed has vertices A(2,1), B(8,1), and C(2,5). They want a second congruent bed made by rotating the herb bed 90° clockwise about point P(4,3), then translating it 3 m left and 2 m down. A circular compost bin is centred at (0,-1) with radius 3 m. The committee requires at least 1 m of clear space between the new bed and the bin. A volunteer claims the transformation preserves area and orientation, so the new bed will be congruent to the original. Before approving the plan, the committee needs the image coordinates, a check for overlap with the bin, and a recommendation if the plan fails. They will use the same soil depth and edging for both beds.
1. Find the coordinates of the image triangle after the full transformation. Show the rule you use.
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Use coordinates. A 90° clockwise rotation about (4,3) maps (x,y) to (4+(y-3), 3-(x-4)) = (y+1, 7-x). Then translating 3 m left and 2 m down gives (x,y) → (y-2, 5-x). Apply to vertices: A(2,1)→(-1,3); B(8,1)→(-1,-3); C(2,5)→(3,3). The image triangle has vertices A'(-1,3), B'(-1,-3), C'(3,3).
2. Verify whether the image triangle is congruent to the original and compute its area. Does the transformation preserve orientation? Explain.
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Original side lengths: AB=6, AC=4, BC=sqrt(6^2+4^2)=sqrt52≈7.21. Image side lengths: A'B'=6, A'C'=4, B'C'=sqrt52. Since all corresponding sides are equal, the triangles are congruent. Area of original = (1/2)(6)(4)=12 m²; image area also 12 m². The volunteer is correct that area is preserved. Orientation is also preserved: a rotation and a translation are both orientation-preserving transformations, so the image is not reflected. The claim about congruence and orientation is valid.
3. The committee’s clearance requirement is 1 m from the bin. Using the image triangle, determine whether the plan meets this requirement. If not, find the smallest whole-number horizontal translation to the right that would create at least 1 m of clearance, keeping all other parts of the transformation.
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The unshifted image triangle has vertices (-1,3), (-1,-3), (3,3). The bin centre (0,-1) lies inside this triangle: at x=0 the triangle spans y from -1.5 to 3, and -1 is in that interval. So the bin overlaps the bed; clearance is not 1 m, and the plan fails. For a rightward shift of k metres, vertices become (k-1,3), (k-1,-3), (k+3,3). The closest point to (0,-1) is on the vertical side x=k-1 at y=-1, distance k-1. For k=4, distance is 3 m, so clearance to the 3 m radius bin is 0 m. For k=5, distance is 4 m, so clearance is 4-3=1 m. The smallest whole-number shift is therefore 5 m to the right.