1. Two different points on the line y=2 are exactly 13 units from the point (7,14). Draw a picture of this situation, and then find the coorfinates of these points.

I drew a line from Point A (7,14) that intersects with y=2 which is Point D. This is one side of the triangle, that is 12 units long. From the equation, you know the length from point A to point C is 13 units, which is a second side of the triangle and the hypotenuse of the triangle. In order to find the base of the triangle, use the pythagorean theorem.

## Table of Contents

## Julianne H. Portfolio Problem

## Christmas Scene

Created with GeoGebra

## 1. Problem 1 on Page 2

1. Two different points on the line y=2 are exactly 13 units from the point (7,14). Draw a picture of this situation, and then find the coorfinates of these points.

I drew a line from Point A (7,14) that intersects with y=2 which is Point D. This is one side of the triangle, that is 12 units long. From the equation, you know the length from point A to point C is 13 units, which is a second side of the triangle and the hypotenuse of the triangle. In order to find the base of the triangle, use the pythagorean theorem.

## Perpendicular lines applet

## Applet 11/5

## Applet 11/12

## Applet 11/22

## Applet 12/3

## applet

Created with GeoGebra

Original line equation:-3x+2y=0 or y=3/2x

Perpendicular line equation:2x+3y=13 or y=-2/3x+13/3

1. Solve system of equation:

3/2x=-2/3x+13/3

13/6x=13/3

x=78/39 =2

2. My vector goes from point C(5,1) to the intersection point(2,3). Therefore, my vector is [-3,2].

3. To find the other vector, double the original vector.

2 x [-3,2]=[-6,4]

4. To find the reflected point, add the original point(5,1) to the vector.

(5,1)+[-3,2]=[2,3]

## Screen cast using geogebra 3d

## Winter break screen cast

## Parallelogram Screen Cast

## Page 32 Problem #8

## Spiderman Problem