C'mon, ! Algebra 1 again! It just looks like a geometry problem! You know, find the equation of a line through two points, where one happens to be the vertex of a triangle and the other happens to be the midpoint of the opposite side. Then repeat that for the other two vertices. Then take two of these line equations and solve the system to find the coordinates of where they intersect, then show this intersection point satisfies the third equation. You then conclude that the medians are concurrent at the centroid.

Algebra 1 is all it is...

Problem 3

Problem 4

Do NOT mess around with this applet until you have tried this problem on your own.

And even then, what is happening in this applet? How can you possible tell from the coordinate rules?

Problem 5

Let's use our newly found DOT PRODUCT TEST!!

I am looking for a point on the x-axis such that the vector draw from (x,0) to (0,12) and the vector drawn from (x,0) to (25,12) will be perpendicular.

The vector from (x,0) to (0,12) is [-x,12]

The vector from (x,0) to (25,12) is [25-x,12]

Then, according to problem 7 on page 22, if they are perpendicular, then

So the points are (16,0) and (9,0). We can confirm this on the applet below.

Problem 6

Use this applet to confirm your answers.

Problem 7

This is a very old puzzle. Think it through.

Problem 8

(a) k=8 make the second vector just twice as long as the first.

(b) We will use our DOT PRODUCT VECTOR TEST:

(4)(k)+(-3)(-6)=0
4k+18=0
4k=-18
k=-4.5

Problem 9

Again, play around with this problem BEFORE you play with the applet.

Problem 10

The first thing that popped onto my mind was a rectangle. Four right angles, but no all sides equal.

Problem 11

Before you play with the applet, think about this...

The point (4,3) is on the line, which means the vector [4,3] is a direction vector. This vector has a length of 5 units. Therefore,

is one point. The other is found by multiplying the coordinates by -1.

Page 24

Problem 1

(a) Graph the points and you will find them to be 10 units apart. If it takes the object 5 seconds to cover this distance, the speed of the object is 2 units/second

(b) Find the equation of the line containing the segment using slope-intercept form. The intercept will be the number I am looking for.

(c) If the object takes 5 seconds to make the trip, then after three seconds, the object will be 3/5 of the way from (-3,1) to (5,7).

Problem 2

Unfold the box, and look for the shortest path on this unfolded box (called a net). There is more than one way to unfold the box. The shortest distance is NOT 42 feet, but is actually 40 feet. How can this be???

Problem 3

What happens when you apply the transformation to the points (0,0) and (0,1)? Calculate the result BEFORE you mess with this applet.

Problem 4

Build a box around the parallelogram, and proceed as with the triangle. Notice, there is a great deal of symmetry.

Problem 5

Let (0,y) be the desired point.

Problem 6

There is only one place for C to go if the order of the vertices is ABCD.

## Table of Contents

## Page 23

## Problem 1

brought to you by Livescribe

## Problem 2

C'mon, ! Algebra 1 again! It just looks like a geometry problem! You know, find the equation of a line through two points, where one happens to be the vertex of a triangle and the other happens to be the midpoint of the opposite side. Then repeat that for the other two vertices. Then take two of these line equations and solve the system to find the coordinates of where they intersect, then show this intersection point satisfies the third equation. You then conclude that the medians are concurrent at the centroid.Algebra 1 is all it is...

## Problem 3

## Problem 4

Do NOT mess around with this applet until you have tried this problem on your own.

And even then, what is happening in this applet? How can you possible tell from the coordinate rules?

## Problem 5

Let's use our newly found

DOT PRODUCT TEST!!I am looking for a point on the

x-axis such that the vector draw from (x,0) to (0,12) and the vector drawn from (x,0) to (25,12) will be perpendicular.The vector from (

x,0) to (0,12) is [-x,12]The vector from (

x,0) to (25,12) is [25-x,12]Then, according to problem 7 on page 22, if they are perpendicular, then

So the points are (16,0) and (9,0). We can confirm this on the applet below.

## Problem 6

Use this applet to confirm your answers.

## Problem 7

This is a very old puzzle. Think it through.

## Problem 8

(a) k=8 make the second vector just twice as long as the first.

(b) We will use our DOT PRODUCT VECTOR TEST:

(4)(k)+(-3)(-6)=0

4k+18=0

4k=-18

k=-4.5

## Problem 9

Again, play around with this problem BEFORE you play with the applet.

## Problem 10

The first thing that popped onto my mind was a rectangle. Four right angles, but no all sides equal.

## Problem 11

Before you play with the applet, think about this...

The point (4,3) is on the line, which means the vector [4,3] is a direction vector. This vector has a length of 5 units. Therefore,

is one point. The other is found by multiplying the coordinates by -1.

## Page 24

## Problem 1

(a) Graph the points and you will find them to be 10 units apart. If it takes the object 5 seconds to cover this distance, the speed of the object is 2 units/second

(b) Find the equation of the line containing the segment using slope-intercept form. The intercept will be the number I am looking for.

(c) If the object takes 5 seconds to make the trip, then after three seconds, the object will be 3/5 of the way from (-3,1) to (5,7).

## Problem 2

Unfold the box, and look for the shortest path on this unfolded box (called a net). There is more than one way to unfold the box. The shortest distance is NOT 42 feet, but is actually 40 feet. How can this be???

## Problem 3

What happens when you apply the transformation to the points (0,0) and (0,1)? Calculate the result BEFORE you mess with this applet.

## Problem 4

Build a box around the parallelogram, and proceed as with the triangle. Notice, there is a great deal of symmetry.

## Problem 5

Let (0,y) be the desired point.

## Problem 6

There is only one place for C to go if the order of the vertices is ABCD.

## Problem 7

[3, -4]. This is a no brainer.

## Problem 8

brought to you by Livescribe

## Problem 9

brought to you by Livescribe

## Problem 10 and 11

If I have done 10 correctly, the point in question is the origin. I think we have addressed these two problems already.