Voronoi diagrams: exam review Exam review One question at a time

Three health clinics are at \(A(1, 2), B(9, 4), C(4, 9)\), where the units are kilometres. Each home goes to its nearest clinic, and the Voronoi diagram shows the area served by each clinic. A home is at \(H(6, 5)\). [6]

1. (a) Find the distance from H to each clinic. [3]

   • M1 for using the distance formula
   • A1 for any two distances correct
   • A1 for all three: A: \(\sqrt{34}=5.83\), B: \(\sqrt{10}=3.16\), C: \(\sqrt{20}=4.47\)

2. (b) Write down the clinic that is nearest to H. [1]

   • A1 for B (the smallest distance, 3.16)

3. (c) A second home G is the same distance from clinic A and from clinic B, and these are its two nearest clinics. Say where G is on the Voronoi diagram. [1]

   • A1 for on the edge (boundary) between the cells of A and B

4. (d) Give one thing that the Voronoi model leaves out. [1]

   • R1 for one valid thing left out, for example roads and travel time (the model uses only the straight-line distance)

A Voronoi diagram has the sites \(P(-2, 3)\) and \(Q(4, 7)\). The line segment RS has equation \(3x+2y=17\) and midpoint \((3, 4)\). [6]

1. (a) Find the midpoint of PQ. [1]

   • A1 for \((1, 5)\)

2. (b) Find the equation of the perpendicular bisector of PQ, in the form \(y=mx+c\). [3]

   • M1 for the gradient of PQ: \(\frac{2}{3}\)
   • M1 for the perpendicular gradient: \(-\frac{3}{2}\), and using the midpoint \((1, 5)\)
   • A1 for \(y=-\frac{3}{2}x+\frac{13}{2}\)

3. (c) Find the equation of the perpendicular bisector of RS, in the form \(px+qy=r\), where p, q and r are integers. [2]

   • M1 for the gradient of RS: \(-\frac{3}{2}\), so the perpendicular gradient is \(\frac{2}{3}\), through \((3, 4)\)
   • A1 for \(2x-3y=-6\)

A city has three fire stations, A, B and C, where the units are kilometres. The diagram shows the Voronoi diagram for the three sites \(A(2, 2), B(8, 4), C(4, 9)\). The point P\((5, 6)\) is marked. [6]

1. (a) Write down the number of edges in the diagram, and the name of the point where they meet. [2]

   • A1 for 3 edges
   • A1 for vertex

2. (b) Find the equation of the edge between A and B, in the form \(px+qy=r\). [3]

   • M1 for the midpoint of AB: \((5, 3)\)
   • M1 for the gradient of AB: \(\frac{1}{3}\), so the edge has gradient \(-3\)
   • A1 for \(3x+y=18\)

3. (c) Write down the site that is nearest to P, and give a reason. [1]

   • A1 for C, because P is inside the cell of C (or: C has the smallest distance, 3.16)

A farmer has soil-moisture sensors across a field. The diagram shows the Voronoi diagram for the six sensors A to F, with the soil moisture at each site. One unit on each axis is 10 m. The points \(X(3, 5)\) and \(Y(3, 3)\) are marked. [7]

1. (a) Use nearest neighbour interpolation to estimate the soil moisture at X. [1]

   • A1 for 19% (X is in the cell of F)

2. (b) The soil moisture at X was later measured as 24%. Find the percentage error in the estimate. [2]

   • M1 for \(\frac{|19-24|}{24}\times 100\)
   • A1 for 20.8%

3. (c) The point Y is close to an edge. Find the distance from Y to F and to A, and hence estimate the soil moisture at Y. [3]

   • M1 for using the distance formula from Y to both sites
   • A1 for \(\sqrt{4}=2.00\) and \(\sqrt{5}=2.24\)
   • A1 for 19% (F is nearer)

4. (d) Explain why the estimate at Y is less reliable than the estimate at X. [1]

   • R1 for saying Y is almost equidistant from two sites (close to an edge), so the other site's value could be as good, whereas X is well inside one cell

Two schools are at \(A(2, 9)\) and \(B\), where the units are kilometres. The perpendicular bisector of AB has equation \(3x-2y=1\). [8]

1. (a) Find the equation of the line AB, in the form \(y=mx+c\). [2]

   • M1 for the gradient of AB, perpendicular to the bisector: the bisector has gradient \(\frac{3}{2}\), so AB has gradient \(-\frac{2}{3}\), and AB passes through A
   • A1 for \(y=-\frac{2}{3}x+\frac{31}{3}\)

2. (b) Use your calculator to find the coordinates of the midpoint M of AB. [2]

   • M1 for solving the equation of AB and the equation of the bisector together (From the calculator)
   • A1 for \((5, 7)\)

3. (c) Hence find the coordinates of B. [1]

   • A1 for \((8, 5)\) (M is the midpoint of AB, so B = 2M - A)

4. (d) A straight road has equation \(y=x+1\). A bus stop on the road is the same distance from the two schools. Find the coordinates of the bus stop. [2]

   • M1 for solving the equation of the bisector and the equation of the road together (From the calculator)
   • A1 for \((3, 4)\)

5. (e) Find the distance from the bus stop to A, in kilometres, to 3 significant figures. [1]

   • A1 for \(\sqrt{26}=5.1\) km (equal to the distance to B)

Four weather stations are at \(A(8, 4), B(6, 2), C(4, 4), D(2, 4)\), where one unit is 1 km. The diagram shows the Voronoi diagram for the four stations. A new station E is built at \(E(6, 3)\), and the Voronoi diagram is redrawn. [10]

1. (a) Find the equation of the perpendicular bisector of E and C, in the form \(y=mx+c\). [3]

   • M1 for the midpoint of E and C: \((5, \frac{7}{2})\)
   • M1 for the gradient of EC: \(-\frac{1}{2}\), so the bisector has gradient \(2\)
   • A1 for \(y=2x-\frac{13}{2}\)

2. (b) The perpendicular bisector of E and A has equation \(4x+2y=35\). Use your calculator to find the coordinates of the vertex where this bisector meets the bisector from part (a). [2]

   • M1 for solving \(y=2x-\frac{13}{2}\) and \(4x+2y=35\) together (From the calculator)
   • A1 for \((6, \frac{11}{2})\)

3. (c) The old diagram has a vertex at \(V(6, 4)\). Show that V is not a vertex of the new diagram. [3]

   • M1 for the distance from V to a site it was equidistant from: \(VA=\sqrt{4}=2.00\) (the same for B, C)
   • M1 for the distance from V to E: \(VE=\sqrt{1}=1.00\)
   • R1 for VE is smaller, so V is inside the cell of E and is not a vertex of the new diagram

4. (d) Write down the stations whose cells share an edge with E, and the number of edges of the cell of E. [2]

   • A1 for A, B, C
   • A1 for 3 edges (one with each of A, B, C)

An island has five villages at \(A(3, 2), B(3, 6), C(8, 3), D(5, 4), E(4, 1)\), where one unit is 1 km. The diagram shows their Voronoi diagram in a rectangular region \(0\le x\le 9\), \(0\le y\le 7\). The vertices inside the region are P\((3, 4)\), Q\((\frac{9}{2}, \frac{5}{2})\), R\((6, 2)\). A waste site is to be placed anywhere in the region, including its boundary, as far as possible from every village. [10]

1. (a) Find the distance from each of P, Q and R to its nearest village. Give your answers to 3 significant figures. [3]

   • M1 for using the distance formula from a vertex to a site next to it
   • A1 for two distances correct
   • A1 for all three: P\((3, 4)\): nearest A\((3, 2)\), \(d=\sqrt{4}=2.00\); Q\((\frac{9}{2}, \frac{5}{2})\): nearest A\((3, 2)\), \(d=\sqrt{2.5}=1.58\); R\((6, 2)\): nearest C\((8, 3)\), \(d=\sqrt{5}=2.24\)

2. (b) Write down which of P, Q and R is farthest from its nearest village. [1]

   • A1 for R, the greatest (2.24)

3. (c) Find the distance from each corner of the region to its nearest village, to 3 significant figures. [3]

   • M1 for the distance from a corner to its nearest site
   • A1 for two correct
   • A1 for all four: (0, 0): nearest A, \(\sqrt{13}=3.61\); (9, 0): nearest C, \(\sqrt{10}=3.16\); (9, 7): nearest C, \(\sqrt{17}=4.12\); (0, 7): nearest B, \(\sqrt{10}=3.16\)

4. (d) Hence write down where the waste site should be placed, and its distance from the nearest village. [2]

   • A1 for the corner \((9, 7)\) with distance 4.12
   • R1 for the corner is farther than every vertex (the greatest vertex distance is 2.24), because the region has a boundary

5. (e) A rule says the waste site must be at least 4 units from every village. Can the rule be met? Give a reason. [1]

   • A1 for yes: the best place is 4.12 units from the nearest village, which is at least 4

Three schools are at \(A(5, 9), B(7, 3), C(5, 5)\), where one unit is 10 km. A waste site is to be placed as far as possible from all three schools. Assume that it goes at the vertex V of their Voronoi diagram. [10]

1. (a) Find the equation of the perpendicular bisector of A and B, in the form \(y=mx+c\). [3]

   • M1 for the midpoint of AB: \((6, 6)\)
   • M1 for the gradient of AB: \(-3\), so the bisector has gradient \(\frac{1}{3}\)
   • A1 for \(y=\frac{1}{3}x+4\)

2. (b) Find the equation of the perpendicular bisector of B and C, in the form \(y=mx+c\). [2]

   • M1 for the midpoint of BC: \((6, 4)\) and the perpendicular gradient \(1\)
   • A1 for \(y=x-2\)

3. (c) Use your calculator to find the coordinates of V. [2]

   • M1 for solving \(y=\frac{1}{3}x+4\) and \(y=x-2\) together (From the calculator)
   • A1 for \((9, 7)\)

4. (d) Show that V is the same distance from A, from B and from C. [2]

   • M1 for the distance formula from V to at least two sites: \(VA^2=20\), \(VB^2=20\), \(VC^2=20\)
   • R1 for all three equal (\(\sqrt{20}\)), so V is the same distance from A, B and C

5. (e) Find the distance from the waste site to the nearest school, in kilometres, to 3 significant figures. [1]

   • A1 for \(\sqrt{20}\times 10=44.7\) km

A country park is a rectangle 12 km by 8 km. Four picnic shelters are at \(A(3, 3), B(5, 7), C(7, 7)\) and \(E(5, 6)\), where one unit is 1 km. The diagram shows the Voronoi diagram for the four shelters. The cell of E is a triangle with vertices \(P(1, \frac{13}{2})\), \(Q(6, \frac{13}{2})\) and \(R(\frac{17}{2}, \frac{3}{2})\). [10]

1. (a) Find the area of the cell of E. [2]

   • M1 for the base and the height of the triangle (or the coordinates in the area formula)
   • A1 for \(\frac{25}{2}\) km\(^2\)

2. (b) Find the percentage of the park that is nearer to E than to any other shelter, to 3 significant figures. [2]

   • M1 for \(\frac{25}{2}\div(12\times8)\times100\)
   • A1 for 13.0%

3. (c) Shelter E is now removed. Find the equation of the perpendicular bisector of A and B, in the form \(y=mx+c\). [3]

   • M1 for the midpoint of A and B: \((4, 5)\)
   • M1 for the gradient of AB: \(2\), so the bisector has gradient \(-\frac{1}{2}\)
   • A1 for \(y=-\frac{1}{2}x+7\)

4. (d) The perpendicular bisector of B and C is \(x=6\). Use your calculator to find the coordinates of the new vertex where the cells of A, B and C meet. [2]

   • M1 for solving the two bisectors together (From the calculator)
   • A1 for \((6, 4)\)

5. (e) Write down the number of vertices of the old diagram that are removed. [1]

   • A1 for 3 (the three vertices of the cell of E)