Category: Geogebra

Posted in Algebra, Geogebra

Solving systems of linear equations by elimination method

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This short investigation  about the graphs of the sum and difference of two or more linear equations may be used as an introductory activity to the lesson on solving systems of linear equations by elimination. It will provide a visual explanation why the method of elimination works, why it’s ok to add and subtract the equations.

The  investigation may be introduced using the GeoGebra applet below.

1. Check the box to show the graph when equations b and c are added.

2. Where do you think will the graph of b – c pass? Check box to verify prediction.

3. Check the box to show graphs of the sum or difference of two equations. What do you notice about the lines? Can you explain this?
[iframe https://math4teaching.com/wp-content/uploads/2011/07/solving_systems_by_elimination.html 700 400]

When equations b and c intersect at A. The graph of their sum will also intersect point A.

b:  x + 2y =1

c:  x – y =-5

a:  2x+y=-4

After this you can then ask the students to think of a pair of equation that intersect at a point and then investigate graph of the sum and difference of these equations. It would be great if they have a graphing calculator or better a computer where they can use GeoGebra or similar software. In this investigation, the students will discover that the graphs of the sum and difference of two linear equations intersecting at (p,r) also pass through (p,r). Challenge the students to prove it algebraically.

If ax+by=c and dx+ey=f intersect at (p,r),

show that (a+d)x+(b+e)y=f +c also intersect the two lines at (p,r).

The proof is straightforward so my advise is not to give in to the temptation of doing it for the students. After all they’re the ones who should be learning how to prove. Just make sure that they understand that if a line passes through a point, then the coordinates of that points satisfies the equation of the line. That is if ax+by=c passes through (p,r), then ap+br=c.


The investigation should be extended to see the effect of multiplying the linear equation by a constant to the graph of the equation or to start with systems of equations which have no solution. Don’t forget to relate the results of these investigations when you introduce the method of solving systems of equation by elimination. Of course the ideal scenario is for students to come up with the method of solving systems by elimination after doing the investigations.

You can give Adding Equations for assessment.

Posted in Geogebra, Geometry

Constructing polygons with equal area

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The power of GeoGebra lies not only for demonstrating a concept but more so for creating a situation where students are made to think, solve problems, and reason mathematically. Here is a sample lesson on how this can be done. The lesson involves the concept of area of triangles and constructing parallel lines applied to problems involving preserving areas of polygons. The most ideal situation is for students to explore on the applets individually or in small group. If this is not possible and the students have no computers, the applets can be projected. The teacher can then call a students or two to explore the applets. The idea is to stimulate students’ thinking to think of explaining why the transformations of the shapes are possible. The students should have a triangle rulers or straight edge with them. The lesson will not be complete if the students cannot devise a procedure for transforming a triangle into other polygonal shapes with the area preserved.

The following applet may be used in the introductory activity to teach about triangles with equal areas. Click here to explore. The applet shows that all triangles with equal base and have equal heights or altitude have equal areas. That’s pretty obvious of course.

The next applet is more challenging but uses the same principle as the first.
Depending on the previous knowledge of your students you can give this second applet right away without showing the first applet on triangles. Click here or the worksheet below explore and answer the problem.

Extension problems using the same principle about area of triangles.

triangle rulers
  1. Given a triangle, construct the following polygons equal in area to the given triangle using triangle rulers and pencil only:  a) parallelogram; b)trapezoid; c) hexagon. (There’s nothing that should prevent you from using GeoGebra to construct them.)
  2. The Rosales and Ronda families are not very happy with their piece of land because of the narrow corners. Help this family to draw a new boundary line without changing the land area of each family.

I will deprive you of the fun if I will show the answers here right away, won’t I?

I will appreciate feedback on this lesson.

Posted in Geogebra, Geometry

Geometric relations – angles made by transversal

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Geometry is a natural area of mathematics for which students should develop reasoning and justification skills and their appreciation of the logico-deductive part of mathematics that build across the grades. Learning tasks therefore should be so designed so that the focus of the learning is on the development of these skills as well and not merely on the learning of facts.

Consider the GeoGebra applets in Figures 1 and 2 below. Which of them will you use for teaching the relationships among the angles made by transversal with parallel lines? Before this lesson of course, the students already learned about linear pairs. Click the figures below to explore the applets before you continue reading.

In the first figure, dragging D or F along the parallel lines, the students will observe that there are angles that will always be equal. Thus from this, they can make the following conjectures:

(1) the alternate interior angles are equal;

(2) the vertical angles are equal;

(3) the corresponding angles (a pair of interior and exterior angles on the same side of the transversal) are equal; and,

(4) the pair of exterior and interior angles on the same side of the transversal sum up to 180 degrees.

In all these cases, the students are reasoning inductively. They will generalize from the measures they observed. Because of this, there seem to be no need for proof since there were bases for the generalizations. The measures of the angles. In this activity students will have learned geometric facts but not the geometric reasoning. Inductive reasoning maybe, but not deductive reasoning.

Contrast the first applet  to the second one. Dragging D or F along the parallel lines, the students will observe that the sum of the pair of exterior -interior angles on the same side of the transversal is always 180 degrees. They will also observe that the other angles also changes. The teacher can then challenge the students to make predictions about the measures of these angles and the relationships among them. These will create a need for proof.

And how should the proof look like? My suggestion is not to be very formal about it like using a two-column proof. For example, to prove that measures of vertical angles are always equal they can set up their proof like these:

To prove p = t:

p + s = 180

s + t = 180

p + s = s + t

p = t.

Students can very well set-up an explanation like this. They have seen it when they learned about solving systems of linear equation. What more, it uses the very important property of equality – the transitive property: If a = c, and b = c, then a = b. Great way to link algebra and geometry.

Posted in Geogebra, Geometry

Problem on proving perpendicular segments

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This problem is a model created to solve the problem posed in the lesson Collapsible.

In the figure CF = FB = FE. If C is moved along CB, describe the paths of F and E. Explain or prove that they are so.

This problem can be explored using GeoGebra applet.  Click this link to explore before you read on.

perpendicular segments

One way to prove that FC is a straight line and perpendicular to AC is to show that FC is a part of a right triangle. To do this to let x be the measure of FCB. Because FCB is an isosceles triangle, FBC and CFB is (180-2x).  This implies that EFB is 180-2x being supplementary to CFB thus CFB must be 2x. Triangle EFB is an isosceles triangle so FBC must be (180-2x)/2. Adding CFB and FBC we have x+ (180-2x)/2 which simplifies to 90. Thus, EB is perpendicular to CB.

The path of F of course is circular with FB as radius.