1. Show geometrically that given any two points in the hyperbolic plane there is always a unique line passing through them. § 23.1 Given point A(x1, y1)

Slides:

Advertisements

Similar presentations

Proving the Distance Formula

Advertisements

Co-ordinate geometry Objectives: Students should be able to

Tangency. Lines of Circles EXAMPLE 1 Identify special segments and lines Tell whether the line, ray, or segment is best described as a radius, chord,

Scholar Higher Mathematics Homework Session

APPLICATIONS OF INTEGRATION

1.3 Use Midpoint and Distance Formulas

Sasha Vasserman.  Two triangles are similar if two pairs of corresponding angles are congruent.

Higher Outcome 1 Higher Unit 1 Distance Formula The Midpoint Formula Gradients Collinearity Gradients of Perpendicular.

Using properties of Midsegments Suppose you are given only the three midpoints of the sides of a triangle. Is it possible to draw the original triangle?

Introduction It is not uncommon for people to think of geometric figures, such as triangles and quadrilaterals, to be separate from algebra; however, we.

APP NEW Higher Distance Formula The Midpoint Formula Prior Knowledge Collinearity Gradients of Perpendicular.

Michael Reyes MTED 301 Section 1-2. Subject: Geometry Grade Level:9-10 Lesson: The Distance Formula Objective: California Mathematics Content Standard.

The Pythagorean Theorem and Its Converse

5.4 Medians and Altitudes A median of a triangle is a segment whose endpoints are a vertex and the midpoint of the opposite side. A triangle’s three medians.

Conics, Parametric Equations, and Polar Coordinates Copyright © Cengage Learning. All rights reserved.

Term 3 : Unit 2 Coordinate Geometry

Formulas to recall Slope: Midpoint: Distance: Definitions to recall Midsegment: Line connecting two midpoints Median: Connects a vertex of a triangle.

1 §23.1 Hyperbolic Models 1 Intro to Poincare's Disk Model. A Point – Any interior point of circle C (the ordinary points of H or h-points) Line – Any.

Analytic Geometry Chapter 5.

Section 8.3 Connections Between Algebra & Geometry

Solving Equations. Is a statement that two algebraic expressions are equal. EXAMPLES 3x – 5 = 7, x 2 – x – 6 = 0, and 4x = 4 To solve a equation in x.

1 Preliminaries Precalculus Review I Precalculus Review II

1-1b: The Coordinate Plane - Distance Formula & Pythagorean Theorem

THE DISTANCE FORMULA ALGEBRA 1 CP. WARM UP Can the set of numbers represent the lengths of the sides of a right triangle? 4, 5, 6.

Higher Unit 1 Distance Formula The Midpoint Formula Gradients

8-1, 1-8 Pythagorean Theorem, Distance Formula, Midpoint Formula

 Congruent Segments – › Line segments that have the same length.  Midpoint – › The point that divides a segment into two congruent segments.  Segment.

More About Triangles § 6.1 Medians

Chapter 4 : Similar Triangles Informally, similar triangles can be thought of as having the same shape but different sizes. If you had a picture of a triangle.

Straight Line Applications 1.1

6/4/ : Analyzing Polygons 3.8: Analyzing Polygons with Coordinates G1.1.5: Given a line segment in terms of its endpoints in the coordinate plane,

Conics can be formed by the intersection

EXAMPLE 2 Use the Midsegment Theorem In the kaleidoscope image, AE BE and AD CD. Show that CB DE. SOLUTION Because AE BE and AD CD, E is the midpoint of.

COORDINATE GEOMETRY Summary. Distance between two points. In general, x1x1 x2x2 y1y1 y2y2 A(x 1,y 1 ) B(x 2,y 2 ) Length = x 2 – x 1 Length = y 2 – y.

Higher Outcome 1 Higher Unit 1 Distance Formula The Midpoint Formula Gradients Collinearity Gradients of Perpendicular Lines The Equation of a Straight.

Proportions and Similar Triangles

Right Triangles Consider the following right triangle.

Geometry Chapter 13 Review. The distance d between points and is: Example 2 Find the distance between (–3, 4) and (1, –4). Why? Let’s try an example to.

M(G&M)–10–9 Solves problems on and off the coordinate plane involving distance, midpoint, perpendicular and parallel lines, or slope GSE: M(G&M)–10–2 Makes.

The Distance Formula (and mid point). What is to be learned? How to calculate the distance between two points.

Week 4 Functions and Graphs. Objectives At the end of this session, you will be able to: Define and compute slope of a line. Write the point-slope equation.

Geometry: Points, Lines, Planes, and Angles

Section 1-1 Points and Lines. Each point in the plane can be associated with an ordered pair of numbers, called the coordinates of the point. Each ordered.

7.1 Apply the Pythagorean Theorem.  Use the Pythagorean Theorem  Recognize Pythagorean Triples.

13.1 The Distance Formula. This should not be new Y X I II III IV Origin.

Introduction The graph of an equation in x and y is the set of all points (x, y) in a coordinate plane that satisfy the equation. Some equations have graphs.

13.1 The Distance Formulas. Review of Graphs Coordinate Plane.

Coordinate Geometry. Coordinate Plane The coordinate plane is a basic concept for coordinate geometry. It describes a two-dimensional plane in terms of.

Coordinate Geometry Please choose a question to attempt from the following:

DAY 1 DISTANCE ON THE PLANE – PART I: DISTANCE FROM THE ORIGIN MPM 2D Coordinates and Geometry: Where Shapes Meet Symbols.

1 Then the lengths of the legs of ABC are: AC = |4 – (–3)| = |7| = 7 BC = |6 – 2| = |4| = 4 To find the distance between points A and B, draw a right triangle.

Problems All Star Team : Marcus Dennis, Wyatt Dillon, Will Clansky.

Chapter 10 Pythagorean Theorem. hypotenuse Leg C – 88 In a right triangle, if a and b are the lengths of the legs and c is the length of the hypotenuse,

HIGHER MATHEMATICS Unit 1 - Outcome 1 The Straight Line.

Geometry Unit 3rd Prep Ali Adel.

13.1 The Distance Formulas.

Midpoint and Distance in the Coordinate Plane

5.4 Midsegment Theorem Geometry 2011.

Right Triangle The sides that form the right angle are called the legs. The side opposite the right angle is called the hypotenuse.

Drawing a sketch is always worth the time and effort involved

1. Graph A (–2, 3) and B (1, 0). 2. Find CD. 8 –2

Introduction It is not uncommon for people to think of geometric figures, such as triangles and quadrilaterals, to be separate from algebra; however, we.

Pythagorean Theorem and Distance

Coordinate Geometry – Outcomes

Distance y x C D A B Lengths parallel to the axes are calculated as if it was a number line. Lengths are always positive. Examples 1) Calculate the length.

Solving Problems Involving Lines and Points

1-6 Midpoint & Distance in the Coordinate Plane

5.7: THE PYTHAGOREAN THEOREM (REVIEW) AND DISTANCE FORMULA

Pythagorean Theorem a²+ b²=c².

Geometry Equations of Circles.

Presentation transcript:

1. Show geometrically that given any two points in the hyperbolic plane there is always a unique line passing through them. § 23.1 Given point A(x1, y1) and B(x2, y2). These points can always be substituted into the general half-circle with center on the x-axis. x 2 + y 2 + ax = b The resulting two equations (one from each point) can be solved except in the case where x1 = x2. However, in this case, the line is a vertical ray with the equation x = x1.

2. Given two points A and B on the hyperbolic with point C between them that the betweenness property holds. I.e. verify that if A – B – C then AC + CB = AB. Note there are two cases for the two “different” types of lines in this model Case 1 – points on a ray: A (a, m), B (b, m) and C (c, m) Case 2 – points on a semicircle:

3. Let A = (1, 3) B = (1, 6), and C = (5, 7) a. Find the equation of the line AB. b. Find the equation of the line AC a. X coordinates are the same. It is a vertical line with equation x = 1. b. Substitute (1, 3) and (5, 7) into x 2 + y 2 + ax = b and solve the resulting two equations 10 + a = b and a = b for a and b yielding a = - 16 and b = - 6. hence the equation of the line AC is x 2 + y 2 – 16 x = - 6

4. Let A = (3, 1), B = (3, 10), C = (12, 20) and D = (24, 16) a. Find the distance from A to B. b. Find the distance from C to D. a. It is a ray so distance AB = 2.30 b.Not a ray so distance CD = 0.69 You need to find M and N first. Use the method of problem 3 to find the equation of the line CD. x 2 + y x = 256. this circle has x-intercepts at M = 32 and N = - 8.

5. Find the angle between the two h-lines x 2 + y 2 = 25 and x 2 + y 2 – 10x = - 9 Solve the two equations simultaneously to get the point of intersection at I(3.4, ). I like to use the derivative of the lines to get the slopes at the point of intersection. First Line: dy/dx = - x/y = = m 1 Second Line : dy/dx = (5 – x)/y = = m 2 Tan  = (m 2 – m 1 )/(1 + m 1 m 2 ) = /.5953 = And  = You may also make an accurate drawing and use a protractor to measure the angle.

6. Show that the two lines x = 5 and x 2 + y 2 – 6x + 5 = 0 are parallel. a. Substituting x = 5 into the second equations yields a y value of 0. this means that both lines have the point (5, 0) in common. This point is on the x-axis and is an ideal point (at infinity) and the two lines are parallel.

7. Let A = (2, 1), B = (2, 3), C = (2, 14) and D = (2, 16). Segments AB and CD have the same Euclidean length. Find the hyperbolic lengths of the two segments and compare. AB = 1.10 and CD = What is going on with distance in this “hyperbolic” model?

8. Given the hyperbolic triangle defined by the points A = (7, 8) B = (1, 20), and C = (1, 10). a. Calculate AC. b. Calculate BC c. Verify that  ABC is an isosceles right triangle. d. Verify that the point D(-24, 0) is the center of the semicircle AB, and use this to calculate each of the angles of  ABC using the slope formula for tan . What is the angle sum of the triangle? a. AC = b. BC = c. See parts a and b. I will confirm it is a right triangle in part d. d.The equation for the three sides are: AB: x 2 + y x = 449 AC: x 2 + y 2 - 2x = 99 and BC: x = 1 Note that the center AB is (- 24, 0). It is always (-a/2, 0) The center of AC is (1, 0) and thus AC and BC form a right angle at vertex C. continued Consider making a drawing of this triangle.

8. Given the hyperbolic triangle defined by the points A = (7, 8) B = (1, 20), and C = (1, 10). d. Verify that the point D(-24, 0) is the center of the semicircle AB, and use this to calculate each of the angles of  ABC using the slope formula for tan . What is the angle sum of the triangle? d.The equation for the three sides are: AB: x 2 + y x = 449 AC: x 2 + y 2 - 2x = 99 and BC: x = 1  C = 90. I will use the method of problem 5 to find the other two angles. To find  A use sides AB and AC. First Line, AB: dy/dx = x + 24/- y = = m 1 at point A Second Line, AC : dy/dx = (1 – x)/y = = m 2 at point A Tan  = (m 2 – m 1 )/(1 + m 1 m 2 ) = / = And  =  A = continued

8. Given the hyperbolic triangle defined by the points A = (7, 8) B = (1, 20), and C = (1, 10). d. Verify that the point D(-24, 0) is the center of the semicircle AB, and use this to calculate each of the angles of  ABC using the slope formula for tan . What is the angle sum of the triangle? d.The equation for the three sides are: AB: x 2 + y x = 449 AC: x 2 + y 2 - 2x = 99 and BC: x = 1  C = 90 and  A = To find  B use sides AB and BC. First Line, AB: dy/dx = x + 24/- y = = m 1 at point B Second Line, BC: BC is a vertical line with slope undefined. To find  B we will first find the angle between line AB and a horizontal line (m 2 = 0) at point B.  B will be 90 – the angle we find. Tan  = (m 2 – m 1 )/(1 + m 1 m 2 ) = 1.25 And  =  B = 90 – =

8. Given the hyperbolic triangle defined by the points A = (7, 8) B = (1, 20), and C = (1, 10). d. Verify that the point D(-24, 0) is the center of the semicircle AB, and use this to calculate each of the angles of  ABC using the slope formula for tan . What is the angle sum of the triangle? d.The equation for the three sides are: AB: x 2 + y x = 449 AC: x 2 + y 2 - 2x = 99 and BC: x = 1  C = 90 and  A = and  B = Notice that the angles opposite equal sides are equal. The sum of the angles of  ABC is

9. Using the information from the previous problem does the Pythagorean Theorem hold in Hyperbolic Geometry? BC = a = , AC = b = , and AB = c = a 2 + b 2 = while c 2 = The Pythagorean Theorem does not hold in hyperbolic geometry.

10. Using the information from the previous problem does the following relationship cosh c = cosh a cosh b hold? BC = a = , AC = b = , and AB = c = cosh a cosh b = and cosh c = Other than round off error the relationship holds. Strange “Pythagorean Theorem”!!!!

11. Find the two h-lines parallel to x 2 + y 2 = 25 through the point (5, 10). The line x 2 + y 2 = 25 has center at the origin and radius 5. thus it crosses the x- axis at (-5, 0) and (5, 0). Each of lines we need will go through one of these points and the given point (5, 10). Consider making a drawing of this triangle. First line. Through (5, 10) and (-5, 0). Using the method of problem 3 substituting into x 2 + y 2 + ax = b gives a = b and 25 – 5a = b which solve giving a = - 10 and b = 75 So the lines is x 2 + y x = 75 Second line. Through (5, 10) and (5, 0). Notice that the x values of these two points are the same so it is a vertical ray with equation x = 5.