Using GeoGebra has several advantages:

・ Ease of use: GeoGebra is a simple and easy to use interface.

・ Multifunctionality: GeoGebra is a multi-functional software package.

・ Free software: that is to say, that the user can download and use for free and “open source” that the user has the right to make modifications.

・ You can find many helps on the software site:

・ http://www.geogebra.org/cms/index.php?option=com_content&task=blogcategory&id=75&Itemid=61

・ A handling document: http://www.geogebra.org/help/geogebraquikstat_fr.pdf

・ The input bar: it allows to enter commands and thus directly define objects; for example: f ( x ) = x 4 − 2 x 2 Figure 1.

・ The toolbar: there are the main construction tools (create point, line, perpendicular line, polygon, circle, angle, etc.); for example: point on object D, extrema A, B and C, the point E in the middle of [BD], in red, the geometric locus of E when the point D varies Figure 1.

・ The algebra window: there is information on the objects created (coordinates, equations, length, etc.); for example: the function f.

・ The worksheet: The worksheet: objects are created and represented graphically, for example the function f and its graph.

・ The spreadsheet allows calculations, statistics and the construction of diagrams and graphs.

The work area of the GeoGebra window shows the geometric aspect of an object and the algebra area shows its analytical aspect, which makes it easier for students to acquire a double perception of a problem.

Thus, the cursor tool allows to create variables or parameters, to deal with a mathematical problem or a problem from everyday life [3].

・ Construction Protocol:

We launch GeoGebra software and we build the following objects: point A, intersection of the coordinate axes, point B, point of the abscissa axis, a point I, point of the abscissa axis, the line (D) which passes through the point I and perpendicular to the abscissa axis, a point C of the straight line (D), the triangle ABC, the orthocenter H of the triangle ABC. Using the command: geometrical locus, we obtain he geometrical locus L of the point H when the point C varies on the line (D) we find: L = (D) Figure 2. We move the point I along the x-axis, we always find L = (D).

If we move point C by the mouse, point H describes the line (D), point H is not defined when A, B and C are aligned.

When we move point I along the abscissa axis, we always find the geometrical locus of H is the line (D).

・ Conclusion: in this case, the geometrical locus of the points H is the line (D).

・ Construction Protocol:

We launch the GeoGebra software and we build the following objects: point A, intersection of the coordinate axes, point B, point of the abscissa axis, a point J, point of the Y axis, the line (D) which passes through the point J and perpendicular to the Y axis, a point C of the straight line (D), the triangle ABC

the orthocenter H of the triangle ABC. Using the command: geometrical locus, we obtain the geometrical locus L of the point H when the point C varies on the line (D) Figure 3 or Figure 4.

We move point J along the ordinate axis to see the shape of the geometrical locus relative to the position of the line (D).

We deduce that:

・ If the ordinate of point J is positive, the geometrical locus is concave, Figure 3.

・ If the ordinate of point J is zero, point H is not defined

・ If the ordinate of point J is negative, the geometrical locus is convex, Figure 4.

Note: We have the impression that the geometrical locus is a parabola.

In this case, the line (D) is oblique, it intersects the abscissa axis at a point I, and it makes an angle α with the abscissa axis; α varies from 0˚ to 180˚.

・ Construction Protocol

We launch the GeoGebra software and we build the following objects: point A, intersection of the coordinate axes, point B, point of the abscissa axis, a point I, point of the x-axis, the angle α by a cursor, the line (D) which makes the angle α with the x-axis passing through the point I, a point C of the line (D), the triangle ABC, the orthocenter H of the triangle ABC, the geometrical locus of point H when point C varies. We represent the angle α by a cursor, point I is displaced along the abscissa axis. we have captured Figures 5-14:

“Five figures when the slope of the line is positive and Five other figures when the slope of the line is negative”.

・ line (D) has a positive slope:

・ The line (D) has a negative slope.

Remarks and conclusions:

・ If I = A, the geometrical locus is the line perpendicular to (D) passing through B, Figure 6 and Figure 11.

・ If I = B, the geometrical locus is the line perpendicular to (D) passing through A, Figure 8 and Figure 13.

・ If I ≠ A and I ≠ B

1) The geometric locus are graphic representations of the numerical functions, these functions are defined continuous and differentiable on the set of real numbers IR except in a, where a is the abscissa of point I

2) If the slope of the line (D) is positive and if I belongs to the segment [AB], the geometrical locus is decreasing, Figure 7.

3) If the slope of the line (D) is positive and if I belongs to the half-line (.,A), or (B,..), the geometrical locus presents a minimum then a maximum, Figure 5 and Figure 9.

4) If the slope of the line (D) is negative and if I belongs to the segment [AB], the geometrical locus is increasing Figure 12.

5) If the slope of the line (D) is negative and if I belongs to the half-line (..,A), or (B,..), the geometrical locus presents a maximum then a minimum, Figure 10 and Figure 14.

6) All the curves pass through points A and B, in fact: the vertical passing through point A intersects the line (D) at a point C, so A is the orthocenter of the triangle ABC, similarly for point B.

Definition: A geometrical locus is a set of points satisfying the conditions, given by a mathematical problem.

By the introduction the Cartesian coordinate system, the points are characterized by their coordinates and the sets of points are characterized by Cartesian equations. Geometric questions are translated algebraically, which sometimes facilitates solutions.

René Descartes (1596-1650) is the first to introduce two-dimensional Cartesian landmarks [7] [8] and Lagrange (1736-1813) is the first to use the equations of lines and planes [9]. Monge (1746-1818) invented the concept of vectors [9].

We present here two methods, which are often used to determine a geometrical place.

The translation method

It consists of:

・ Choose an orthonormal coordinate system, so that the data have simple equations.

・ Graph the data.

・ Translate data and conditions by algebraic relationships.

・ We obtain a relation, or an equation verified by the coordinates of the points of the geometrical locus.

・ We study the converse and we trace the place in question.

The generator method:

The generatrices are curves, which contain the point P whose geometrical locus is sought. The method consists of:

・ Choose a reference, it is not necessarily orthonormal.

・ Graph the data.

・ Specify the equations of the fixed lines and the coordinates of the fixed points.

・ Identify the generators.

・ Find the equations of the generators.

・ Eliminate the parameter between the two generators, thus obtaining the equation of the geometrical locus.

・ Analyze and build the geometric locus.

In 1945 George Polya published his famous book How to Solve It; more than a million copies have been sold; this book has been translated into 17 languages. In this book, George Polya identifies four basic principles for maximum luck in solving a problem, which are [10]:

・ First Principle: Understand the problem.

・ Second Principle: Devise a plan.

・ Third Principle: Carry out the plan.

・ Fourth Principle: Look back.

But there is problem and problem, for our problem, we will follow other steps without forgetting the advice of George Polya.

Formulation of the Problem:

Problem: “In the plane, ABC is a triangle, of orthocenter H. Determine the geometrical locus L of the points H when C moves on a fixed line (D)”.

We consider an orthonormal coordinate system, ( A , A B → , A B ′ → ) .

Let H be the orthocenter of the triangle ABC, its coordinates are ( r , s ) .

Mathematization of the Problem

・ First case: (D) is perpendicular to the abscissa axis at a point I ( a , 0 ) .

The line (D) is a height, so x = a is the abscissa of H, its ordinate varies from −∞ to +∞ when the point C varies along the line (D), therefore (D) is the geometrical locus of H, Figure 2.

・ Second case: (D) is perpendicular to the ordinate axis at a point J ( 0 , a ) .

The equation of the line (D) is y = a , C is a point on the line (D), its coordinates are ( r , a ) with r is a real varies. Let H be the orthocenter of the triangle ABC, its coordinates are ( r , a ) . Point H checks:

B H → ⋅ A C → = 0 (1)

So:

( r − 1 ) ( r − 0 ) + ( s − 0 ) ( a − 0 ) = 0 (2)

So

g ( x ) = x ( 1 − x ) / a (3)

The Equation (3) is the equation of a parable of vertex E ( 1 / 2 , 1 / 4 a ) and axis of symmetry x = a , Figure 3 and Figure 4.

・ Third case (D) is an oblique line.

Of coordinate system, the line (D) is oblique, so it intersects the abscissa axis at a point I ( a , 0 ) where a is a real number. The equation of the line (D) is written

y = ( x − a ) tg ( α ) . (4)

With α is an element of the interval ] 0 , π / 2 [ ∪ ] π / 2 , π [ .

C is a point on the line (D), its coordinates are ( r , ( r − a ) tg ( α ) ) with r is a real number.

Let H be the orthocenter of the triangle ABC, its coordinates are ( r , s )

Mathematization of the Problem

Point H checks

B H → ⋅ A C → = 0 (5)

Using Cartesian coordinates, we find the coordinates (r, s) of the orthocenter H verify:

r ( r − 1 ) + s ( r − a ) tg ( α ) = 0 (6)

Resolution of problem:

・ First case: if a = 0 , the Equation (6) becomes

r ( r − 1 + s tg ( α ) ) = 0 . (7)

Or, r varies, so

s = ( 1 − r ) / tg ( α ) . (8)

That we write

y = ( 1 − x ) / tg ( α ) . (9)

So, the geometrical locus is a line which passes through point B, Figure 6 and Figure 11.

・ Second case: if a = 1 , the Equation (6) becomes:

( r − 1 ) ( r + s tg ( α ) ) = 0 . (10)

Or, r varies, so

s = − r / tg ( α ) . (11)

That we write

y = − x / tg ( α ) . (12)

It is a line, which passes through point A, Figure 8 and Figure 13.

Third case: a ≠ 0 and a ≠ 1 , The numbers a and α are parameters. α ≠ 0 because ABC is an unfastened triangle. So:

s = r ( 1 − r ) / ( r − a ) tg ( α ) . (13)

Hence, the geometric locus of the orthocenter H is the curve C ( a , α ) representing the function

g ( a , α ) ( x ) = x ( 1 − x ) / ( x − a ) tg ( α ) . (14)

Note that the curves C ( a , α ) and C ( a , π − α ) are symmetrical with respect to the abscissa axis because tg ( α ) = − tg ( π − α ) , also note that: M ( x , y ) is a point of C ( a , π / 4 ) equivalent to M ′ ( x , y / tg ( α ) ) is a point of C ( a , α ) because tg ( π / 4 ) = 1 .

We are only interested in the representation of the curve C ( a , π / 4 )

g ( a , π / 4 ) ( x ) = x ( 1 − x ) / ( x − a ) (15)

The function g ( a , π / 4 ) is a defined, continuous and differentiable on ] − ∞ , a [ ∪ ] a , + ∞ [ and its derivative is:

g ′ ( a , π / 4 ) ( x ) = ( − x 2 + 2 a x − a ) / ( x − a ) 2 (16)

g ′ ( a , π / 4 ) ( x ) = 0 if and only if the discriminant a 2 − a is positive, its roots are x 2 = − a − a 2 − a and x 2 = − a + a 2 − a . ( a 2 − a < 0 if 0 < a < 1 Otherwise a 2 − a > 0 ).

If 0 < a < 1 , g ′ ( a , π / 4 ) ( x ) < 0 , then the function g ( a , π / 4 ) is strictly decreasing.

The table of variation (Table 1 and Table 2):

The line x = a is an asymptote; to find the oblique asymptotes, we change the variable: x = X + a we find: x ( 1 − x ) / ( x − a ) = − x − a + 1 + ( a − a 2 ) / ( x − a ) , We deduce that the line y = − x − a + 1 is an asymptote which is below the curve C ( a , π / 4 ) on ] − ∞ , a [ and above the curve C ( a , π / 4 ) on ] a , + ∞ [ .

Graphical representation using GeoGebra software:

We launch the GeoGebra software. The results were shown as Figures 15-17.

We define the cursor, we enter the function g ( a , π / 4 ) ( x ) and the equations of the asymptotes the line y = − x − a + 1 and the line x = a .

IF a < 0 or a > 1 , g ′ ( a , π / 4 ) vanishes at x 1 and x 2 moreover it is positive on ] x 1 , x 2 [ because x 1 < a < x 2 and ( − x 2 + 2 a x − a ) is positive for x = a .