00:01
In this problem, our job is to find the location of the school, given the fact that it must be equidistant from the two towns.
00:09
So one town is at the point a, 2 comma negative 2, and the other town is at the point b, 8, 5.
00:16
And we're given that the school must lie on the straight line given by the equation minus x plus 7y equals negative 4.
00:25
So we can go ahead and solve this equation for y, get a sort of a standard equation for the line.
00:33
On which the school lies.
00:36
So we see that 7y is x minus 4, and so we know that that line has equation y equals 1 7th x minus 4 7th.
00:49
So let's sketch a graph of this line along with the two towns.
00:55
That will give us a better picture of what we're trying to find.
00:58
We'll make some room for that sketch.
01:11
So we need to go out to at least 8 here on our x -axis.
01:21
We will set up our axes accordingly.
01:35
So let's plot our two towns, a and b.
01:40
So here is the town at point a, and here is the town at point b.
01:53
Let's graph our line.
01:54
We notice that it has an x intercept of 4 .0.
02:00
We can check some other points on this line.
02:04
We know when x is 4 .y is 0.
02:07
When x is 11, y is 1.
02:10
1.
02:11
X is negative 3.
02:12
Y is negative 1.
02:14
So we have those points on our graph as well.
02:21
And so the tail must, excuse me, the new school must lie on this line.
02:26
Let's see if we can connect those points a little bit better.
02:28
I'm going to go back and start over.
02:29
Here we go.
02:31
If we can connect those.
02:35
This line has a slope, as you can see, very close to 0.
02:38
It's 1 7th.
02:39
So i'll draw that in there.
02:42
So we're looking for a point on this line, roughly about right here.
02:46
We call that point c.
02:48
With the distance from a to c equals the distance from b to c.
02:53
And it has to be on that line.
02:56
Well, one way to find that point is to go back and find the perpendicular bisector of the segment from a to b.
03:02
So if we connect a to b here, we find this perpendicular bisector of that segment.
03:09
That line must go through c.
03:12
So if we do that, we'll have to find the midpoint of a and b, which is going to be here.
03:19
We'll do that calculation just a minute, confirm that.
03:23
We want the perpendicular bisector here that's going to come through and intersect the school's line, the line in blue, at the point c.
03:32
So there's our setup using our diagram...