Here are two screen shots from a calculator page on my Nspire. I tried to solve the equation in two different ways. What did I do to solve each equation in each screen shot? What are the solutions?

What does this screen shot say about the solution?

Problem 2 and 3

This problem is getting at the idea that when you reflect a point over a line, the line becomes the perpendicular bisector of the point and its reflection. For which of the two points given is it easier to find the reflection?

Problem 4

There are examples of this shape in my room in the back. Pick one up and COUNT!

Problem 5

If you draw the diagonals of a rhombus, they are perpendicular and they bisect each other, and they divide the rhombus into four congruent right triangles.

Problem 6

If point C is 8 units from A, then point C is 8/13 ths of the way from A to B. You have solved this problem a gazillion times.

Problem 7

I am looking for points on the y-axis that satisfy the conditions of the problem. Therefore, the points I am looking for have the form (0,y).

Squaring both sides, we have

Apply the quadratic formula to this, we have

Problem 8

The strategy (for now) is to construct the smallest possible rectangle around the triangle, find the area of that, then subtract off the area of the right triangles around the outside of the triangle.

For the figure below, this results in

Very soon, we will learn a couple of other methods, one involving vectors, and another known as Heron's Formula.

Problem 9

You should double-click on the applet to explore how it was made.

Problem 10

Look back at Page 18 and modify the proof shown there. You should then create an illustration in GeoGebra and include this proof on your page.

Problem 11

Graph the points as they suggest, and calculate the lengths of AB and AC. Use trig to find the measures of the angles they ask you to find. You should include the general statement of the theorem (called the Isosceles Triangle Theorem) and YOUR PROOF of the statement on your page.

Problem 12

Prove it, then write up the proof on your page.

Page 22

Problem 1

At each step, you need to be asking yourself "WHY?"

So the width of the uniform border is either 24 yards or 4 yards. I can disregard the 24 yard answer. The border is 4 yards wide.

Problem 2

Problem 3

From playing with the applet below, it certainly LOOKS like an isometry. In fact, a 90 degrees counter clockwise rotation.

To show that T(x,y)=(-y,x) is an isometry, I need to take two arbitrary points, say (a,b) and (c,d), and I need to show that the distance between these two points is equal to the distance found between, when I run these points through the transformation, their images (-b,a) and (-d,c).

In other words, I want to show

Well, we can ignore the radicals, and try to convince ourselves that

I can obviously cancel some things, so I am left to show that

Expand both sides, and you will find that they are identical. This means that the transformation is an isometry.

Problems 4 and 5

Just to get my bearing, I sketch a picture of each triangle.

For the figure on the left, the angles at vertex B and C are both 62 degrees.

For the figure on the right, the angle a C is 56, but the angle at A is 68 degrees.

Problem 6

THIS IS A MAJOR RESULT! BIG BIG BIG IDEA!

If we assume that vector [c,d] is perpendicular to [a,b], then thinking about slopes, the vector [c,d] must be a scalar multiple of [-b,a] (in other words, perpendicular, but not necessarily the same length), so [c,d] = k[-b,a] for some k.

Turning the idea in Problem 6 around. I hope you agree that if I were to sketch the vector [a,b] on graph paper, it would have a slope of

Put in another way, if I were to multiply the component a by the slope of the vector, I would get the component b.

Now, if it is true that ac + bd = 0, then is must be true that ac = -bd.

Looking carefully at this, this says the component d can be found by multiplying the component c by the PERPENDICULAR SLOPE of the vector [a,b]. Hence, the vector [c,d] must be perpendicular to [a,b] whenever ac + bd = 0

Problem 8

Is it possible to argue that one triangle is just a scale copy of the other?

Problem 9

There are a number of ways to approach this problem, and over the next few pages, we will develop a vector approach to calculating the area, but for the time being, you will need to calculate the area of the large rectangle bounding the triangle, then subtracting off the area of the triangles and small rectangle outside the triangle.

Problem 10

OK HONORS Geometry students...this is an Algebra 1 question disguised as a geometry question. You know, finding the equation of three lines (which are altitudes), each passing through three different points (which just happen to be the vertices of a triangle), and each perpendicular to three different lines (where these three different lines just happen to be the sides of a triangle). Then you want to take two of these equations you wrote and you want to solve the system to find their point of intersection. Then you want to take this point of intersection and show that it also satisfies the the third equation. Hence, you conclude that the three altitudes are concurrent.

## Table of Contents

## Page 21

## Problem 1

Here are two screen shots from a calculator page on my Nspire. I tried to solve the equation in two different ways. What did I do to solve each equation in each screen shot? What are the solutions?

What does

thisscreen shot say about the solution?## Problem 2 and 3

This problem is getting at the idea that when you reflect a point over a line, the line becomes the perpendicular bisector of the point and its reflection. For which of the two points given is it easier to find the reflection?

## Problem 4

There are examples of this shape in my room in the back. Pick one up and COUNT!

## Problem 5

If you draw the diagonals of a rhombus, they are perpendicular and they bisect each other, and they divide the rhombus into four congruent right triangles.

## Problem 6

If point C is 8 units from A, then point C is 8/13 ths of the way from A to B. You have solved this problem a gazillion times.

## Problem 7

I am looking for points on the y-axis that satisfy the conditions of the problem. Therefore, the points I am looking for have the form (0,y).

Squaring both sides, we have

Apply the quadratic formula to this, we have

## Problem 8

The strategy (for now) is to construct the smallest possible rectangle around the triangle, find the area of that, then subtract off the area of the right triangles around the outside of the triangle.

For the figure below, this results in

Very soon, we will learn a couple of other methods, one involving vectors, and another known as Heron's Formula.

## Problem 9

You should double-click on the applet to explore how it was made.

## Problem 10

Look back at Page 18 and modify the proof shown there.

You should then create an illustration in GeoGebra and include this proof on your page.## Problem 11

Graph the points as they suggest, and calculate the lengths of

ABandAC. Use trig to find the measures of the angles they ask you to find.You should include the general statement of the theorem (called the Isosceles Triangle Theorem) and YOUR PROOF of the statement on your page.## Problem 12

Prove it, then

write up the proof on your page.## Page 22

## Problem 1

At each step, you need to be asking yourself "WHY?"

So the width of the uniform border is either 24 yards or 4 yards. I can disregard the 24 yard answer. The border is 4 yards wide.

## Problem 2

## Problem 3

From playing with the applet below, it certainly

LOOKSlike an isometry. In fact, a 90 degrees counter clockwise rotation.To show that T(x,y)=(-y,x) is an isometry, I need to take two arbitrary points, say (a,b) and (c,d), and I need to show that the distance between these two points is equal to the distance found between, when I run these points through the transformation, their images (-b,a) and (-d,c).

In other words, I want to show

Well, we can ignore the radicals, and try to convince ourselves that

I can obviously cancel some things, so I am left to show that

Expand both sides, and you will find that they are identical. This means that the transformation is an isometry.

## Problems 4 and 5

Just to get my bearing, I sketch a picture of each triangle.

For the figure on the left, the angles at vertex B and C are both 62 degrees.

For the figure on the right, the angle a C is 56, but the angle at A is 68 degrees.

## Problem 6

THIS IS A MAJOR RESULT! BIG BIG BIG IDEA!If we assume that vector [c,d] is perpendicular to [a,b], then thinking about slopes, the vector [c,d] must be a scalar multiple of [-b,a] (in other words, perpendicular, but not necessarily the same length), so [c,d] = k[-b,a] for some k.

Therefore, ac + bd = a(-kb) + b(ka) = -kab + kab = 0.

BE CERTAIN YOU FOLLOW THIS REASONING!!## Problem 7

BE CERTAIN YOU FOLLOW THIS REASONING!!Turning the idea in Problem 6 around. I hope you agree that if I were to sketch the vector [

a,b] on graph paper, it would have a slope ofPut in another way, if I were to multiply the component

aby the slope of the vector, I would get the componentb.Now, if it is true that

ac+bd= 0, then is must be true thatac= -bd.Looking carefully at this, this says the component

dcan be found by multiplying the componentcby thePERPENDICULAR SLOPEof the vector [a,b]. Hence, the vector [c,d] must be perpendicular to [a,b] wheneverac+bd= 0## Problem 8

## Problem 9

## Problem 10

OK

HONORSGeometry students...this is an Algebra 1 question disguised as a geometry question. You know, finding the equation of three lines (which are altitudes), each passing through three different points (which just happen to be the vertices of a triangle), and each perpendicular to three different lines (where these three different lines just happen to be the sides of a triangle). Then you want to take two of these equations you wrote and you want to solve the system to find their point of intersection. Then you want to take this point of intersection and show that it also satisfies the the third equation. Hence, you conclude that the three altitudes are concurrent.C'mon...this is

...ALGEBRA 1## Problem 11

Use this applet to check your answers