Page 25


Problem 1


Look back at the last problem on Page 22.

Part a


To find the y-intercept, you substitute in x=0 and solve for y in terms of m.



Likewise, to find the x-intercept, you subsitute y=0 and solve for x in terms of m



Part b


The area of this triangle is



Part c


Graph this function on your Nspire, and find the minimum as usual. It is a nice exercise to think WHY the area must be at least 20.

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)



Problem 2


parallelogram_area.png

You will need to study this carefully, and convince yourself of each statement.

  • First off, the coordinates of R are indeed (a+c,b+d).

Now, we can combine some of the triangular regions to form rectangular regions.

  • Triangle 2 and 5 can be combined to form a rectangle with an area of ab.
  • Triangle 3 and 6 can be combined to form a rectangle with an area of cd.
  • The rectangles 1 and 4 have areas of cb.
  • The area of the large rectangle is (a+c)(b+d).
  • The area of the parallelogram is (a+c)(b+d) - ab - cd - 2cb.

Expanding, this is ab + ad + cb + cd - ab - cd - 2cb = ad - cb.

It is possible that this calculation could result in a negative number, so it should be |ad-cb|.



Problem 3


Page 25 Problem 3
brought to you by Livescribe




Problem 4


This transformation is NOT an isometry. Notice that the segment connecting the points (0,0) and (0,2) is mapped to the segment connecting the points (0,0) and (0,1). These two segments have different lengths, hence, the transformations is not an isometry.



Problem 5


Ahhh...the old EQUIDISTANT PROBLEM. In your head, the two points are





Problem 6


This configuration is two parallel lines: y=3 and y=-3.



Problem 7


Have you seen this before?

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)



Problem 8 and 9



page 25 problem 8
brought to you by Livescribe

page 25 problem 9
brought to you by Livescribe





Problem 10


Draw in any two diagonals. Convince yourself that the two triangles shown are congruent. Based upon that, convince yourself that the two diagonals drawn in are congruent. Why can we conclude that all the diagonals are the same length?

pentagon.png



Problem 11


page25_problem11.png
Page 25 Problem 11
brought to you by Livescribe




Problem 12


Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)
page 25 problem 12
brought to you by Livescribe




Problem 13


I actually constructed the applet below with just the line and the slider controlling the slope m. As I dragged the slider, I tried to visualize where the square would be situated in the first quadrant. I noticed that the line would need to pass through the point (4,4), which is where the upper right corner of the square would be located.

So, the equation of the line is y-2=m(x-5). Substituting y=4 and x=4 and solving for m, we find that m=-2.

The applet below confirms this.

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)



Problem 14


Taking advantage of our new area tool from the top of this page, I draw in two vectors from A as shown below. The vector from A to D is [3,4] and the vector from A to B is [5,1]. Using the area formula we derived, the area is |(3)(1)-(5)(4)| = 17.

page_25_problem_14.png




Page 26


Problem 1


Looking at the picture below, there are two points to find. One is 3/10 of the way from E to F. The other is just the other direction.

In other words, point B is found by



while the other is found by



page_26_problem_1.png



Problem 2 and 3


Been there, done that. You should be able to do all three parts BY HAND, as well as be able to find the equation of the Euler Line.



Problem 4


Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)
page 26 problem 4
brought to you by Livescribe




Problem 5


Using a vector idea, the point (6,9) is one point that would satisfy the conditions of the problem. In fact, and point on the circle centered at the origin and passing through the point (6,9) will work. The equation of this circle is





Problem 6


The x-intercept crossing occurs when y=0, or when -2+4t=0. This means when t=0.5. The x-intercept is (6.5,0).

Likewise, the y-intercept crossing occurs when x=0, or when 5+3t=0/ This means when t=-5/3. The y-intercpet is (0,-26/3)



Problem 7


This is the perpendicular bisector of (3,5) and (7,-1)



Problem 8


It would seem that however you place the triangle, symmetry would need to play a part. But of course, how do we *PROVE* it?

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)



Problem 9


This is a variation of the "find a point some fraction of the way from this point to that point." It is almost obvious by looking at the picture.

page_26_problem_9.png



Problem 10


We did a problem similar to this on page 24, number 8. There are a number of arguments you could make. The easiest one you could make is a symmetry argument, saying that if you draw a line through the midpoint of side AI and point E, segments IF and AD are reflection images.

To make a congruent triangle argument, you could draw in a segment from H to F and from B to D. You could easily show that triangle HGF is congruent to triangle BCD. You could then argue that triangle IHF is congruent to triangle ABD. Then appeal to CPCTC.

You should carefully construct both arguments.

page_26_problem_10.png



Problem 11


Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)



Problem 12


Page 26 problem 8
brought to you by Livescribe




Problem 13


Page 26 Problem 13
brought to you by Livescribe




Problem 14


When you transform something geometrically, representing the transformation works according to Newton's Third Law: Every action has an equal and opposite reaction.

When you slide an object to the right 3 units and up 10 units, you represent this be thinking, "what would it take to move the transformed line BACK to its original position?" Obviously, 3 units left and 10 units down.

This is represented in the equation by replacing x with x-3 and replacing y with y-10. The equation of the transformed line is
5(x-3)+7(y-10)=35, or 5x+7y=120.

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)