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Explanation:
For now, we're only going to consider Newmarket and Markham, which are points N and M respectively.
Let's find the slope of line NM through use of the slope formula.
m = (y2-y1)/(x2-x1)
m = (3-8)/(7-2)
m = -5/5
m = -1
Line NM has a slope of -1. Its perpendicular slope is +1 or just 1.
Rule: If the slopes multiply to -1, then the lines are perpendicular.
Or you can apply the negative reciprocal rule to go from -1 to 1.
We'll keep this perpendicular slope in mind for later.
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Now let's find the midpoint of points N and M.
Add up the x coordinates and divide by 2: (x1+x2)/2 = (2+7)/2 = 4.5
Do the same for the y coordinates: (y1+y2)/2 = (8+3)/2 = 5.5
The midpoint of N and M is at (4.5, 5.5)
The perpendicular line must go through these two points. I'm going to skip a few steps here, but basically we go from y = mx+b to y = x+1 after plugging in those x,y values and the perpendicular m we found earlier.
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In short, the perpendicular bisector for line segment NM is the equation y = x+1.
If you follow the same steps as the previous two sections, but you apply them for points M and V instead, then you should find that the perpendicular bisector is y = -3x+9. I'm skipping steps here as well.
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At this point, we have the system of equations
Let's use substitution to solve for x and y
y = x+1
-3x+9 = x+1
-3x-x = 1-9
-4x = -8
x = -8/(-4)
x = 2
y = x+1
y = 2+1
y = 3
So together, x = 2 and y = 3 pair up to get the ordered pair solution (x,y) = (2,3)
This is the location where the theme park should be placed. Let's say that C = (2,3). I'm using C for center.
You can use the distance formula to confirm that segments CN, CM and CV are all the same length, which would verify that point C is equidistant from all the other points. Furthermore, it means that points N, M and V are all on the same circle centered at C.
I know it says not to use a graph, but it still helps to see what's going on. Also, it's a handy way to confirm the answer. I used GeoGebra to make the graph. See the diagram below.
Side note: you only need 2 perpendicular bisectors to find point C. You could use three, but that's overkill since they all intersect at the same location.
