Detailed Answer

a) The sites are the points around which the Voronoi diagram is made.

b) The vertexes are where the edges of the diagram intersect.

c) The cells are drawn by the edges. On this graph, we only have three, when we should have four.

d) The equation of the line that goes through \(D\) and \(C\) is \(y = 3\). (We see it graphically, but we can also check that the gradient is \(\frac{y_C - y_D}{x_C - x_D} = \frac{5 - 5}{7 - 3} = 0\) and the \(y\)-intercept is \(y_D - 0 \cdot x_D = 3\).)

This means that the perpendicular bisector will be perpendicular to the \(x\)-axis, and have an equation of the form \(x = k\).

The midpoint \(I\) of the segment \([DC]\) has coordinates \(x_I = \frac{x_D + x_C}{2} = \frac{3 + 7}{2} = 5, \ y_I = \frac{y_D + y_C}{2} = \frac{3 + 3}{2} = 3\), which gives \((5, 3)\)

So the equation is \(x = 5\).

e) The bisector is drawn in red, the edge in green.

f) We name the points in red on the following diagram \(M\), \(N\), \(O\), \(P\), \(Q\), and \(R\), to make the explanations easier.

Line \((OP)\): \( \text{gradient} = \frac{y_N - y_O}{x_N - x_O} = \frac{5 - 10}{5 - 0} = -1\); \(y\)-intercept: can be read on the graph.

Line \((NM)\): \(y = 5\)

Line \((MP)\): \( \text{gradient} = \frac{y_P - y_M}{x_P - x_M} = \frac{8 - 5}{13 - 10} = 1\); \(y\)-intercept = \(y_M - 1 \cdot x_M = 5 - 10 = -5\)

Line \((QM)\): \( \text{gradient} = \frac{y_M - y_Q}{x_M - x_Q} = \frac{5 - 0}{10 - 9} = 5\); \(y\)-intercept = \(y_Q - 5 \cdot x_Q = 0 - 5 \cdot 9 = -45\)

Detailed Answer

a) Line (AB) is parallel to the \(y\)-axis, and is the bisector of segment \([T_2 T_3]\). This means line \([T_2 T_3]\) is perpendicular to the \(y\)-axis and has for equation \(y = 8\). The midpoint \(I\) of \([T_2 T_3]\) is both on (AB) and \([T_2 T_3]\), so its coordinates are \(x_1 = 1\), \(y_1 = 8\).

We already know \(T_2\) must be on \(y = 8\), and we can find its \(x\)-coordinate by using \(I\), as \(I\) is equidistant to \(T_2\) and \(T_3\). The distance \(IT_3\) is:

\[ \sqrt{(4 - 1)^2 + (8 - 8)^2} = 3 \]

For the distance \(IT_2\) to be 3, then we need \(x_{T_2} = -2\).

We can check by calculating:

\[\sqrt{(1 - x_{T_2})^2 + (8 - y_{T_2})^2} = 3 \]

Replacing \(y_{T_2} = 8\), this gives us:

\[ \sqrt{(1 - x_{T_2})^2} = 3 \implies (1 - x_{T_2})^2 = 9 \implies (1 - x_{T_2}) = 3 \text{ or } -3 \implies x_{T_2} = -2 \text{ or } 4 \]

b) The midpoint of \([T_3 T_4]\) has coordinates \(\left(\frac{4 + 6}{2}, \frac{8 + 2}{2}\right) = (5, 5)\).

The gradient of \([T_3 T_4]\) is \(\frac{2 - 8}{6 - 4} = \frac{-6}{2} = -3\). The gradient of the perpendicular bisector will be \(\frac{1}{3}\), since \(-3 \times \frac{1}{3} = -1\).

To find the intercept, we use the coordinates (5, 5).

\[ \frac{1}{3} \times 5 + \text{y-intercept} = 5 \implies \frac{5}{3} + \text{y-intercept} = 5 \implies \text{y-intercept} = 5 - \frac{5}{3} = \frac{10}{3} \]

Thus, the equation of the edge is \(y = \frac{1}{3}x + \frac{10}{3}\)

c) Matthew has the \(x\)-coordinate of \(T_1\) and the \(y\)-coordinate of \(T_4\). Drawing the two lines shows that Matthew is on the red left cell, so Matt gets the best signal from \(T_2\).

Detailed Answer

a) Each of the four cells covers some part of the island!

b) Point (6, 10) is within the blue cell, which is around the site of tower \(A\).

c) This is the point with coordinates (6, 7) which is in the green cell, covered by tower \(B\).

d) Island 3 is only covered by tower \(C\), as it is entirely within the orange cell.

e) Going on a straight line between these two points, we intersect the edge between the orange and yellow cells at coordinates (12, 9).

Detailed Answer

a) She’s within the cell around the site for site \(C\), which means she is closest to \(C\).

b) Being on the vertex between the three sites \(A\), \(B\), and \(C\), Simon is at equal distance from all three.

c) Since it makes no difference for Simon, they can meet at coffee shop \(C\).

d) Simon is still at equal distance from coffee shops \(A\) and \(B\). Continuing the perpendicular bisector of segment \([AB]\), we can see that point \(R\) is located underneath it, meaning it is closer to \(A\) than \(B\). So if \(C\) is closed, they should go to \(A\).

Detailed Answer

a) We can see they're at the intersections of edges.

b) Since the bus stop is on the vertex between the sites \(B\), \(C\), and \(D\), these are Lisa's options.

c) That distance is calculated by \[ \sqrt{(x_C - x_2)^2 + (y_C - y_2)^2} = \sqrt{(10 - 7)^2 + (5 - 4)^2} = \sqrt{3^2 + 1^2} = 3.2 \text{ km} \]

d) We calculate the coordinates of the middle of \([AC]\): \[ \left(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}\right) = \left(\frac{2 + 10}{2}, \frac{9 + 5}{2}\right) = (6, 7) \] which are the coordinates of Monument B.

e) (i) & (ii) For the distance between monuments \(C\) and \(B\), we calculate the distance with the formula: \[ \sqrt{(10 - 6)^2 + (5 - 7)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \text{ km} \] For the distance between monument \(C\) and \(A\), we know from the previous question that we can simply multiply this by two, and we get 10 km.

f) She first walked 3.2 km in question c), then 10 km in question e). We only have to calculate the distance between monument \(A\) and Bus Stop 3: \[ \sqrt{(3 - 2)^2 + (6 - 9)^2} = \sqrt{1^2 + 3^2} = \sqrt{4} = 2 \text{ km} \] So in total, she walked 3.2 + 10 + 2 km = 15 km.

Detailed Answer

a) \(\sqrt{(4.5-3)^{2}+(0-1.5)^{2}}=2.1 \mathrm{~km}\)

b) The cell around site \(A\) is a rectangle, the area it covers is Area \(A = 3 \times 2.5 = 7.5 \mathrm{~km}^{2}\).

The cell around site \(B\) is a rectangle and a triangle. \(Area B = 3 \times 1.5 + \frac{1.5 \times 1.5}{2} = 5.6 \mathrm{~km}^{2}\).

The cell around site \(C\) is also a rectangle and triangle. \(Area C = 1.5 \times 2.5 + \frac{1.5 \times 1.5}{2} = 4.9 \mathrm{~km}^{2}\).

So school A covers the highest area in the city.

c) If Kelly lives the same distance away from \(A\) and \(B\), then she lives on the perpendicular bisector of segment \(AB\), which is line \((I_2J)\). Line \((I_2J)\) has the equation \(y = 1.5\), so the \(y\) coordinate of Kelly's house is \(y_K = 1.5\).

d) For the \(x\)-coordinate, we can use the formula for the distance with either of the two points \(A\) or \(B\). Using \(A\), we find:

\[\sqrt{(x_A - x_K)^{2} + (y_A - y_K)^{2}} = \sqrt{(2 - x_K)^{2} + (2.5 - 1.5)^{2}} = \sqrt{(2 - x_K)^{2} + 1^{2}} = 2.3\]

\[\Rightarrow (2 - x_K)^{2} + 1 = 2.3^{2}\]

\[\Rightarrow 2 - x_K = \sqrt{2.3^{2} - 1} \text{ or } -\sqrt{2.3^{2} - 1}\]

\[\Rightarrow x_K = 2 - \sqrt{2.3^{2} - 1} \text{ or } 2 + \sqrt{2.3^{2} - 1}\]

\[\Rightarrow x_K = -0.1 \text{ or } 4.0\]

And from the diagram, we know \(x_K\) cannot be negative, so \(x_K = 4.0\)

e) Kelly lives at coordinates (4,1.5), which is in the cell of site C. This means that Kelly works at school C.

f) \(\sqrt{(x_K - x_C)^{2} + (y_K - y_C)^{2}} = \sqrt{(4 - 4)^{2} + (1.5 - 2.5)^{2}} = \sqrt{(-1)^{2}} = 1 \mathrm{~km}\).

Detailed Answer

a) For each of these, we will be using points with easy-to-read coordinates on the diagram.

Between \(A\) and \(C\): Gradient: \(\frac{5-10}{5-0} = -1\), y-intercept: \(10\), equation: \(y = -x + 10\)

Between \(C\) and \(B\): A horizontal line with intercept \(y = 5\).

Between \(A\) and \(B\): A vertical line with \(x\)-coordinate \(x = 5\).

b) (i) & (ii)

The cell around \(A\) and \(C\) are a square and a triangle together.

\[ Area_A = \text{Area}_C = 5^2 + \frac{5 \times 5}{2} = 37.5 \mathrm{~km}^{2}\]

(iii) The cell around \(B\) is a square.

\[ Area_B = 5^2 = 25 \mathrm{~km}^{2}\]

c) We start with the perpendicular bisector between D and C. It ends at the edge between C and A.

Then we can add the bisector between D and A. It ends on the same point.

Which gives us the new Voronoi diagram:

Detailed Answer

a) It has to be on one of the two vertices between three edges, so either \(E\) or \(F\). We can calculate which one is the furthest away from the nearest sites.

Distance \(CF\): \[ \sqrt{(4.4-7)^2 + (3.1-1)^2} = 3.3 \]

Distance \(ED\): \[ \sqrt{(6-4)^2 + (6-3.7)^2} = 3.0 \]

\(F\) is the vertex furthest from any site, so this is the place we want to build a pond.

b) \(BT\) is a horizontal line, since \(B\) and \(T\) have the same \(y\)-coordinate. The perpendicular bisector will be parallel to the \(y\)-axis. Its \(x\)-coordinate will be the \(x\)-coordinate of the middle of segment \([BT]\), so: \[ \frac{x_B + x_T}{2} = \frac{6+8}{2} = 7 \]

c) Gradient of \(DT\): \[\frac{y_T - y_D}{x_T - x_D} = \frac{2-6}{8-6} = -2 \] So the gradient of a perpendicular line must be \(\frac{1}{2}\), as \(-2 \times \frac{1}{2} = -1\)

Coordinates of the middle of \([DT]\): \[ \left(\frac{x_D + x_T}{2}, \frac{y_D + y_T}{2}\right) = \left(\frac{6+8}{2}, \frac{6+2}{2}\right) = (7,4) \]

This gives us the \(y\)-intercept of the bisector: \[ \frac{1}{2} \times 7 + b = 4 \Rightarrow b = 4 - \frac{7}{2} = 0.5 \]

d) We first draw point \(T\), and the two bisectors from the previous questions.

Then we can draw the two relevant edges.

e) We have to check if the new vertex is closer or further away from the other sites than vertex \(F\), so we calculate its distance with point \(T\): \[ \sqrt{(8-7)^2 + (2-4)^2} = \sqrt{1+5} = \sqrt{6} = 2.4 \]

This vertex is closer to other sites than \(F\), so the pond doesn't need to change positions.

Detailed Answer

a) It must be on one of the vertexes between three edges. For each of them, we'll calculate their distance to their closest sites:

Distance between \((5,4)\) and site 3: \[ \sqrt{(5-9)^{2}+(4-2)^{2}} = \sqrt{(-4)^{2}+2^{2}} = \sqrt{20} = 4.5 \, \mathrm{dam} \]

Distance between \((10,5.7)\) and site 3: \[ \sqrt{(10-9)^{2}+(5.7-2)^{2}} = \sqrt{1^{2}+3.7^{2}} = 3.8 \, \mathrm{dam} \]

So the best place is the vertex at coordinates \((5,4)\).

b) (i) Gradient of \(S_5S_1\):

\[ \frac{4-8}{5-7} = \frac{4}{2} = 2 \] Gradient of the perpendicular bisector: \(-\frac{1}{2}\)

Middle of segment: \[ \left(\frac{5+7}{2} ; \frac{4+8}{2}\right) = (6,6)\]

\(y\)-intercept of the perpendicular bisector: \[ -\frac{1}{2} \times 6 + b = 6 \Rightarrow b = 6 + \frac{6}{2} = 9 \]

(ii) Gradient of \(S_5S_2\): \[\frac{4-2}{5-1} = \frac{2}{4} = \frac{1}{2} \] Gradient of the perpendicular bisector: \(-2\).

Middle of segment: \[ \left(\frac{5+1}{2} ; \frac{4+2}{2}\right) = (3,3) \]

\(y\)-intercept of the perpendicular bisector: \[ b = 3 + 2 \times 3 = 9 \]

(iii) Gradient of \(S_5S_3\): \[ \frac{4-2}{5-9} = \frac{2}{-4} = -\frac{1}{2} \] Gradient of the perpendicular bisector: \(2\).

Middle of segment: \[ \left(\frac{5+9}{2} ; \frac{4+2}{2}\right) = (7,3) \]

\(y\)-intercept of the perpendicular bisector: \[ b = 3 - 2 \times 7 = -11 \]

c) First, we draw the bisectors we previously calculated equations for:

Then we can use these to have the new Voronoi diagram: