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.

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?

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.

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 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

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.

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?

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.

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.

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.

## Table of Contents

## Page 25

## Problem 1

Look back at the last problem on Page 22.

## Part a

To find the

y-intercept, you substitute inx=0 and solve foryin terms ofm.Likewise, to find the

x-intercept, you subsitutey=0 and solve forxin terms ofm## 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.

## Problem 2

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

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

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

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## 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 andy=-3.## Problem 7

Have you seen this before?

## Problem 8 and 9

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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?

## Problem 11

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## Problem 12

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## 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). Substitutingy=4 andx=4 and solving form, we find thatm=-2.The applet below confirms this.

## 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 26

## Problem 1

Looking at the picture below, there are two points to find. One is 3/10 of the way from

EtoF. The other is just the other direction.In other words, point

Bis found bywhile the other is found by

## 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

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 wheny=0, or when -2+4t=0. This means whent=0.5. Thex-intercept is (6.5,0).Likewise, the

y-intercept crossing occurs whenx=0, or when 5+3t=0/ This means whent=-5/3. They-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?

## 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.

## 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.

## Problem 11

## Problem 12

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## Problem 13

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## 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

xwithx-3 and replacingywithy-10. The equation of the transformed line is5(

x-3)+7(y-10)=35, or 5x+7y=120.