Year: 2010

Geonext Tutorial 1 – Constructing an Equilateral Triangle

Geonext, Software Tutorials

Geonext is aJava-written interactive and free geometry software. It is developed by the Lehrstuhl für Mathematik und ihre Didaktik (Chair of mathematics and its didactics) of the University of Bayreuth in Germany and released under the GNU General Public License.

Figure 1 - The Geonext Window.

 1.) Click the New Board button.
2.) To draw a circle with center A and passing through B, click the Circle tool, click the drawing board to determine the center of the circle, then click another location to determine its radius. Notice that Geonext, automatically names the points in alphabetical order.


3.) With the Circle tool still active, click point B and then click point A to create a circle, with center B passing through point A.

Figure 2 - Circles with centers A and B.
4.) Click the Point tool, and click one of the intersections of the two circles. Notice that a point D was also constructed.
5.) Next, we hide the circles and point D, leaving only points A, B and C on the drawing area. To hide the objects, click Objects menu from the menu bar, click Special Properties and click Hide. Click the two circles.

Figure 3 - The circumference of the two circles are hidden.

Notice, that hidden objects are colored pitch. The figure above is shown when you click point D. That means that Geonext is asking you which object to hide. A circle ca, cb or point D. Since both circles are already hidden, you just have to click point D.
6.) Next, we will use the Polygon tool to draw triangle ABC. To draw the triangle, click the Polygon tool, click point A, click point B, click point C and then click point A to close the triangle.
Q1: Move the vertices of the triangle. What do you observe?
7.)Now, to verify that the triangle is equilateral, we can do two things: reveal the measure of the interior angles or the side lengths.  To reveal the side length, click Texts and Calculations, then click Measure Distance, then click the three sides of the triangle.
8.) You can also measure the angles using the angle tool and using three points. For example, if you want to measure angle B, click the Objects menu, click Texts and Calculations, then click Measure Angle, then click point A, click point B and click point C.
9.  Move the vertices of the triangle. What do you observe?
Q2:.  Explain why ABC is always an equilateral triangle.
10.  Click the File menu and click Save if you want to save your file.

There is also a similar construction here using GeoGebra.

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A Closer Look at the Midpoint Formula

High School Algebra, High School Geometry, High School Mathematics

Introduction

If we want to find the midpoint of segment CD in Figure 1, where the coordinates of points C and D are (-2,5) and (-2,1), it is clear that the length of CD is 5 - 1 = 4 units.  To determine the midpoint of CD, we want to get the coordinates of the point which is 2 units away from both points C and D.  Hence, we have to divide 5 - 1 = 4 by 2, and add the result to 1 or subtract the result from 5.  Summarizing, the expression that would describe the value of the y-coordinate of the midpoint would be \displaystyle\frac{(5 - 1)}{2} + 1 or  5 - \displaystyle\frac{(5 - 1)}{2}. This means that the midpoint of CD is (0,2).

If we want to get the midpoint of AB, using the same reasoning above, the expression that would describe the x-coordinate of the midpoint would be \displaystyle\frac{(6 - 3)}{2} + 3 or 6 - \displaystyle\frac{(6 - 3)}{2}. This means that the midpoint of AB is (1.5,0).

Figure 1 – Horizontal line AB and vertical line CD in the coordinate plane.

Generalizing our observation above, if we have a vertical segment with their endpoints having coordinates (a, y_1) and (a,y_2) (see Figure 2), we can get its midpoint  using the following formula \displaystyle\frac{y_2 - y_1}{2} + y_1  or y_2 - \displaystyle\frac{y_2 - y_1}{2} + y_1

Figure 2 – Generalized coordinates of a vertical and a horizontal line.

Simplifying, both the expressions above result to \displaystyle\frac{y_2 + y_1}{2}. Similarly, for a horizontal segment with endpoints having coordinates (x_1, b) and (x_2, b) can be computed by the expression \displaystyle\frac{x_1 + x_2}{2}.

Midpoint of a Slanting Segment

The preceding derivations are only valid for vertical and horizontal segments. In Figure 3, segment AB is neither horizontal nor vertical. To investigate the midpoint of AB, we draw vertical segment PQ coinciding with the y-axis with endpoints having y-coordinates the same as those of the y-coordinates of segment AB (see Figure 3). We also draw a horizontal line RS coinciding the x-axis with endpoints having  x-coordinates the same as those of the x-coordinates of segment AB. Looking at Figure 3, it is clear that the coordinates of the midpoint of the PQ is (0,3).

If we draw a horizontal line from (0,3) towards segment AB (see yellow dashed segment), and draw a vertical line from the intersection M to segment RS, it seems that the intersection of the yellow dashed line and segment RS is (3.5,0) which is the midpoint of RS. From here, it is tempting to ask the following question:

“If the midpoint of PQ is (0,3) and the midpoint of RS is (3.5,0), is the midpoint of AB, (3.5,3)?

Figure 3 – A non-horizontal segment AB with midpoint M.

Generalizing the questions above, we might want to ask “If the midpoint the point with coordinates A (x_1,y_1) and B (x_2, y_2) equal to (\displaystyle\frac {x_1 + x_2}{2}, \displaystyle\frac{y_1 + y_2}{2})?”

If we extend the QA and the horizontal yellow dashed line to the right, we can from two right triangles as shown in Figure 4. From the statements above, we want to show that point M is the midpoint of AB.

To show that M is the midpoint of AB, it is sufficient to show that AM is congruent to MB.  This leads us to Figure 5, where we label the right angles of the two triangles T and U. We will now show that triangle AUM is congruent to triangle MTB. If so, then we can show that AM is congruent to MB since they are the corresponding sides of the given triangles.

Proof that AUM is congruent to MTB

Since PB and MT are horizontal segments, we can consider AB as a transversal of parallel segments PB and the segment containing MT.  It follows that angle TBM and angle UMA are congruent because they are corresponding angles. It is also clear that BT is congruent to MU, and angle T is congruent to angle U since they are both right angles. Therefore, by the ASA congruence theorem, AUM is congruent to MTB.  (For an explanation of parallel lines and transversals, click here).

Since corresponding parts of congruent triangles are congruent, AM is congruent to MB. Hence, M is the midpoint of AB.

Figure 5 – The triangle produced by extending the horizontal lines passing through the three points.

Note that our proof did not talk about coordinates, but the general case.  That is, if the coordinates of A are ( x_1, x_2) and the coordinates of B are (y_1, y_2), the coordinates of M is equal to (\displaystyle\frac{x_1 + x_2}{2},\displaystyle\frac{y_1 +yx_2}{2} ).

Delving Deeper

There are also other ways to show that the midpoint of AB is M.  ( x_1,y_1) and ( x_2,y_2) lies on  the midpoint formula. The details of the solutions are left to the reader as an exercise.

  • From Figure 3, draw two right triangles with hypotenuse AM and hypotenuse AB and show that AM is half of AB.
  • Using the distance formula, show that the distance between point A and point M is the same as the distance between point M and point B.
  • Show that AM and MB has the same slope.
  • Get the equation of the line containing AB, and substitute the coordinates of M to the equation of line AB.
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Free Algebra Ebooks

Resources and Freebies

Below is a collection of free algebra ebooks that are downloadable. They include math ebooks on High School Algebra, Trigonometry, College Algebra, Linear Algebra, and Abstract Algebra.

Elementary Algebra and Trigonometry Ebooks

  1. Advanced Algebra II: Conceptual Explanations by Kenny Felder
  2. Advanced Algebra II: Activities and Homework by Kenny Felder
  3. Algebra I by  CK-12 Foundation
  4. CK-12 Trigonometry by Mara Landers and Brenda Meery
  5. College Algebra by Carl Stitz and Jeff Zeager
  6. Concepts of Algebra by Christopher Cooper
  7. Elementary Mathematics by William Chen and Xuan Duong
  8. Elementary Algebra by Christopher Cooper
  9. Elementary Algebra by Denny Burzynski and Wade Ellis
  10. Elementary Trigonometry by W. E. Paterson
  11. Intermediate Algebra by Department of Mathematics, College of the Redwoods
  12. Intermediate Algebra by John Blakely
  13. Trigonometric Delights by  Eli Maor
  14. Trigonometry by Michael Corral

Abstract and Linear Algebra

  1. A First Course in Linear Algebra, by Rob Beezer.
  2. A Second Semester of Linear Algebra by S. E. Payne
  3. Abstract Algebra by John A. Beachy, William D. Blair
  4. Abstract Algebra On Line by John A. Beachy, William D. Blair
  5. Abstract Algebra: Theory and Applications by Thomas Judson
  6. Abstract Algebra:The Basic Graduate Year by Professor Robert Ash
  7. Algebra: Abstract and Concrete by Frederick M. Goodman
  8. Elementary Linear Algebra, by Keith Matthews.
  9. Elements of Abstract and Linear Algebra by  E.H. Connell
  10. Linear Algebra and Jim Hefferon
  11. Linear Algebra by Macquarie University
  12. Linear Algebra and Matrices by Martin Fluch
  13. Linear Algebra Done Wrong by Sergei Treil
  14. Linear Algebra by Professor Jim Hefferon
  15. Linear Algebra for Informatics and José Figueroa-O’Farrill
  16. Linear Algebra: An Introduction to Linear Algebra for Pre-Calculus Students by Tamara A. Carter, Richard A. Tapia, Anne Papakonstantinou

Other Algebra Books

  1. A Course in Universal Algebra, by Stanley Burris
  2. Lie Algebras by Professor Sternberg

You may also want to check my Free Calculus Ebooks post.

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