The Voronoi diagram below shows four hotels in a small tow
e .5lsg(ri-f, .szoxqn represented by points wi
.xs( qsi-e,r5 gzlf.o th coordinates A(−4,4), B(3,5), C(3,−3), and D(−1,3). The vertices $V_1$, $V_2$ and $V_3$ are also shown. Distances in the direction of the x and y axes are measured in increments of 100 metres.
1.Find the midpoint of AD. M (a,b) a= b=
2.Hence, find the equation of the line that passes through $V_1$ and$V_2$. y =ax+b a = b =
The equation of line that passes through $V_1$ and $V_3$ is y=−2x+6.
3.Find the coordinates of $V_1$.(a,b) a= b=
The coordinates of $V_2$ are (−5,−4) and the coordinates of V3V3 are (2.5,1).
4.Find the distance from $V_1$ to $V_2$. Give your answer to the nearest metre. D ≈
5.Given that the distance from $V_1$ to $V_3$ is 783 metres, find the angle $V_2\widehat{V_1}V_3$. Give your answer to the nearest degree. $V_2\widehat{V_1}V_3$ ≈ $^{\circ}C$
6.Hence, find the area of the Voronoi cell containing hotel D, giving your answer in $m^2$, to three significant figures. The manager of hotel D believes that the larger the area of triangle $V_1V_2V_3$, the more people will stay at hotel D.A ≈ $m^2$
7.State one criticism of the manager's belief.