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    The Voronoi diagram below shows four supermarkets represented by points with coordinates $\mathrm{A}(0,0), \mathrm{B}(6,0), \mathrm{C}(0,6)$ and $\mathrm{D}(2,2)$. The vertices $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ are also shown. All distances are measured in kilometres. The equation of (XY) is $y=2-x$ and the equation of $(\mathrm{YZ})$ is $y=0.5 x+3.5$. The coordinates of Y are $(-1,3)$ and the coordinates of Z are $(7,7)$.

    Question
    SLPaper 2

    The Voronoi diagram below shows four supermarkets represented by points with coordinates A(0,0),B(6,0),C(0,6)\mathrm{A}(0,0), \mathrm{B}(6,0), \mathrm{C}(0,6)A(0,0),B(6,0),C(0,6) and D(2,2)\mathrm{D}(2,2)D(2,2). The vertices X,Y,Z\mathrm{X}, \mathrm{Y}, \mathrm{Z}X,Y,Z are also shown. All distances are measured in kilometres. The equation of (XY) is y=2−xy=2-xy=2−x and the equation of (YZ)(\mathrm{YZ})(YZ) is y=0.5x+3.5y=0.5 x+3.5y=0.5x+3.5. The coordinates of Y are (−1,3)(-1,3)(−1,3) and the coordinates of Z are (7,7)(7,7)(7,7).

    1.

    Find the midpoint of [BD][\mathrm{BD}][BD]

    [2]
    Verified
    Solution

    (2+62,2+02)\left(\frac{2+6}{2}, \frac{2+0}{2}\right)(22+6​,22+0​) (4,1)(4,1)(4,1) Note: Award A0\boldsymbol{A0}A0 if parentheses are omitted in the final answer.

    2.

    Find the equation of (XZ)(\mathrm{XZ})(XZ)

    [3]
    Verified
    Solution

    attempt to substitute values into gradient formula (0−26−2=)−12\left(\frac{0-2}{6-2}=\right)-\frac{1}{2}(6−20−2​=)−21​ therefore the gradient of perpendicular bisector is 2 so y−1=2(x−4)y-1=2(x-4)y−1=2(x−4) (y=2x−7)(y=2x-7)(y=2x−7)

    3.

    Find the coordinates of X

    [3]
    Verified
    Solution

    identifying the correct equations to use: y=2−xy=2-xy=2−x and y=2x−7y=2x-7y=2x−7 evidence of solving their correct equations or of finding intersection point graphically (M1) (3,−1)(3,-1)(3,−1) A1 Note:Accept an answer expressed as "x=3,y=−1x=3, y=-1x=3,y=−1".

    4.

    Determine the exact length of [YZ]

    [2]
    Verified
    Solution

    attempt to use distance formula YZ=(7−(−1))2+(7−3)2YZ=\sqrt{(7-(-1))^{2}+(7-3)^{2}}YZ=(7−(−1))2+(7−3)2​ =80(45)=\sqrt{80}(4\sqrt{5})=80​(45​)

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    5.

    Given that the exact length of [XY][\mathrm{XY}][XY] is 32\sqrt{32}32​, find the size of XY^ZX \hat{Y} ZXY^Z in degrees

    [4]
    Verified
    Solution

    METHOD 1 (cosine rule) length of XZ is 80(45,8.94427…)\sqrt{80}(4\sqrt{5}, 8.94427\ldots)80​(45​,8.94427…) Note: Accept 8.94 and 8.9 attempt to substitute into cosine rule cos⁡XY^Z=80+32−802×8032\cos X\hat{Y}Z=\frac{80+32-80}{2\times\sqrt{80}\sqrt{32}}cosXY^Z=2×80​32​80+32−80​ (=0.316227…)(=0.316227\ldots)(=0.316227…) Note: Award A1\boldsymbol{A1}A1 for correct substitution of XZ, YZ, 32\sqrt{32}32​ values in the cos rule. Exact values do not need to be used in the substitution. (XY^Z=)71.6∘(71.5650…∘)(X\hat{Y}Z=) 71.6^{\circ}(71.5650\ldots^{\circ})(XY^Z=)71.6∘(71.5650…∘) Note: Last A1\boldsymbol{A1}A1 mark may be lost if prematurely rounded values of XZ , YZ and/or XY are used.

    METHOD 2 (splitting isosceles triangle in half) length of XZ is 80(45,8.94427…)\sqrt{80}(4\sqrt{5}, 8.94427\ldots)80​(45​,8.94427…) Note: Accept 8.94 and 8.9. required angle is cos⁡−1(32280)\cos^{-1}\left(\frac{\sqrt{32}}{2\sqrt{80}}\right)cos−1(280​32​​) Note: Award A1 for correct substitution of XZ (or YZ), 322\frac{\sqrt{32}}{2}232​​ values in the cos rule. Exact values do not need to be used in the substitution. (XY^Z=)71.6∘(71.5650∘)(X\hat{Y}Z=) 71.6^{\circ}(71.5650^{\circ})(XY^Z=)71.6∘(71.5650∘) Note: Last A1\boldsymbol{A1}A1 mark may be lost if prematurely rounded values of XZ , YZ and/or XY are used.

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    6.

    Hence find the area of triangle XYZ

    [3]
    Verified
    Solution

    (area=)128032sin⁡71.5650…(\text{area}=)\frac{1}{2}\sqrt{80}\sqrt{32}\sin 71.5650\ldots(area=)21​80​32​sin71.5650… OR (area=)123272(\text{area}=)\frac{1}{2}\sqrt{32}\sqrt{72}(area=)21​32​72​ =24 km2=24 \text{ km}^2=24 km2

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    7.

    State one criticism of this interpretation

    [1]
    Verified
    Solution

    Any sensible answer such as:

    There might be factors other than proximity which influence shopping choices. A larger area does not necessarily result in an increase in population. The supermarkets might be specialized / have a particular clientele who visit even if other shops are closer. Transport links might not be represented by Euclidean distances. etc.

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