Jack_Demo

=media type="custom" key="6549011"= =Trying out the Tools=

On this page, I will be testing some of the tools I might want to use for embedding math content in a Wiki page. For demonstration purposes I'll do a page about perpendicular bisectors in a triangle. In the actual course, this would build on previous skills such as the midpoint formula and finding equations of lines perpendicular to other lines. There would ideally be active pages containing that content, but here we'll just have dead links to nonexistent pages, so don't be too disappointed.

=The Perpendicular Bisector of a Segment=

The **perpendicular bisector** of a line segment is the unique line such that Notice that in this case, the name actually tells you all the important information - it is perpendicular, and it bisects the segment (cuts the segment exactly in half).
 * it is perpendicular to the original line segment
 * it passes through the midpoint of the line segment

Every line segment has exactly one perpendicular bisector.

Another nice way to characterize perpendicular bisectors is the following property:

A point lies on the perpendicular bisector of segment amath bar {A B} endmath if and only if it is equidistant from point A and B.
In the following Sketchpad applet, point M is the midpoint of AB, and line PM is the perpendicular bisector. Notice the circle centered at P, which demonstrates that points A and B are equidistant from P. Try dragging points A and B around a little, and slide point P along the perpendicular bisector.

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The sketch suggests something very interesting about perpendicular bisectors and triangles.

=Perpendicular Bisectors in Triangles=

Suppose we have a triangle ABC. We draw the perpendicular bisectors of sides AB and AC. Now, these perpendicular bisectors must intersect (why?). Let's call their intersection O. Point O is equidistant from A and B, since it is on the perpendicular bisector of AB. Similarly it is equidistant from A and C. Therefore, it must be equidistant from B and C, and must also lie on the perpendicular bisector of BC. We can draw a circle centered at O that passes through all three vertices of the triangle. This circle is called the //circumscribed circle// or //circumcircle// for short. The point O is called the //circumcenter//, since it is the center of the circumcircle. The radius of this circle is, unsurprisingly, called the //circumradius//.

This construction is unique -- there is only one circle passing through all three vertices of any triangle.

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Now let's look at how do calculations about the circumcenter for a triangle in the coordinate plane.

=Example: Triangles in the Coordinate Plane= Suppose that we have a triangle whose vertices are given by points A(0,0), B(12,0), and C(4,4). Let's compute the equations of the pependicular bisectors of this triangle, and intersect them to find the coordinates of the circumcenter.

In order to find the circumcenter, we start by finding the equations of any two perpendicular bisectors. Let's start with the bisector of amath bar { A B } endmath, since it is the easiest. We know that it passes through the midpoint of amath bar { A B } endmath, so we use the midpoint formula to compute the coordinates of the midpoint.( amath (0 + 12)/(2), (0 + 0)/(2) endmath ) = ( amath 6 , 0 endmath).

Now, the perpendicular bisector of AB is a vertical line, since it is perpendicular to the x-axis. The vertical line through (6,0) is just amath x = 6 endmath.

Now, let's do the perpendicular bisector of amath bar { A C } endmath. The midpoint of amath bar { A C } endmath is amath ( (0 + 4)/(2), (0 + 4)/(2) ) = (2 , 2) endmath.

In order to make the line perpendicular to amath bar { A C } endmath, we must use the slope of amath bar {A C } endmath. This slope is amath ( 4 - 0)/(4 - 0) = 1 endmath,. A line perpendicular to that will have slope -1. We now plug the point amath ( x, y ) = ( 2, 2) endmath and the slope amath m = -1 endmath into the formula amath y = m x + b endmath to get the equation of the line. amath 2 = -1 * 2 + b endmath. This gives amath b = 4 endmath, so the equation of the perpendicular bisector is amath y = - x + 4 endmath.

Finally, to get the coordinates of the circumcenter, we can intersect our two perpendicular bisectors, amath x = 6 endmath and amath y = - x + 4 endmath. Plugging 6 in for amath x endmath in the second equation yields amath y = - 6 + 4 = - 2 endmath, so the coordinates of point O are amath ( 6, -2 ) endmath. This matches up well with our sketch, which is always worth checking for reasonableness.