PLANES, LINES AND POINTS A B A A B C A B C m A

A B C D Points A, B and D are collinearPoints A, B and C are not collinear BD BA andare opposite rays B is between A and D C is not between A and D Any two points are collinear. Two points define a line.

M A B C D E

Two lines intersect as a point Two planes intersect at a line

Draw and label one diagram that includes all of the following: a. Plane R, points X, Y and Z coplanar, but not collinear. b. XY and ZX c. AB non-coplanar to plane R, but intersecting plane R at point X R X Y Z A B

1.3 Segments and Their Measures

Ruler Postulate The points on a line can be matched one to one with real numbers. The real number that corresponds to a point is the coordinate of the point. The distance between points A and B, written as AB, is the absolute value of the difference between the coordinates of A and B. 9 A B 3 Find AB: AB= 3-9 or 9-3 AB=6

Segment Addition Postulate: If B is Between A and C, then AB + BC = AC If AB + BC = AC then B is between A and C B A C B A C B is between A and C B is not between A and C

Ex: if DE=2, EF=5, and DE=FG, find FG, DF, DG, & EG. D E F G FG=2DF=7DG=9EG=7

Ex. Y is between X and Z. Find XY, YZ and XZ if: XY= 3x + 4 YZ= 2x + 5 XZ= 9x - 3 (3x + 4) + (2x +5) = 9x - 3 5x + 9 = 9x = 4x x = 3 XY = 3(3) + 4 XY = 13 YZ = 2(3) + 5 YZ = 11 XZ = 9(3) -3 XZ = 24

If two line segments have the same lengths they are said to be congruent. The symbol for congruence is ≅ If AB = 5 and CD = 5 then, AB = CD, distances are equal AB ≅ CD, segments are congruent

The Distance Formula A B If A ( x₁, y₁) and B ( x₂, y₂) are points in a coordinate plane then the distance between A and B is: AB = √ ( x₂ - x₁ )² + ( y₂ - y₁ )²

G H (3, 2) (11, 6) Using the Distance Formula GH = √ (11 -3)² + (6 - 2)² GH = √ 8² + 4² GH = √ GH = √ 80 GH = 4√ 5

The Distance Formula comes from the Pythagorean Theorem a² + b² = c² where a and b are the legs of a right triangle and c is the hypotenuse. AB is the hypotenuse of a right triangle (x₂ - x₁)² = a, the length of the horizontal leg (y₂ - y₁)² = b, the length of the vertical leg A B b a c (x₂, y₂) (x₁, y₁)

G H (3, 2) (11, 6) Using the Pythagorean Theorem (11,2) a c b a² + b² = c² √ a² + b² = c √ ∣ 11-3 ∣² + ∣ 6-2 ∣² = c √ 8² + 4² = c √ = c √ 80 = c 4√5 = c

J (2,5), K (7,11) Find JK a is the distance between 2 and 7, a=5 b is the distance between 5 and 11, b=6 R (5,12), S (9, 2) Find RS a is the distance between 5 and 9, a=4 b in the distance between 12 and 2, b=10 JK = √ 5² + 6² JK = √61 RS = √ 4² + 10² RS = √ 116 RS = 2 √ 29

An angle consists of two different rays that have the same initial point. The rays are the sides of the angle. The initial point is the vertex of the angle. A C B side vertex

Angles are named using three points, the vertex and a point on each ray which makes up the angle. The vertex is always the second point listed. If an angle stands alone it can be named by its vertex only. H J I ∠JHI or ∠IHJ or ∠H

You should not name any of these angles as  H because all three angles have H as their vertex. The name  H would not distinguish one angle from the others. H J I K ∠JHI or ∠IHJ ∠JHK or ∠KHJ ∠IHK or ∠KHI

The measure of ∠ A is denoted by m∠ A. The measure of an angle can be approximated using a protractor, using units called degrees(°). For instance, ∠ BAC has a measure of 50°, which can be written as m∠ BAC = 50°. C A B

Angles that have the same measure are called congruent angles If ∠BAC and ∠DEC both measure 75˚ then m ∠BAC = m∠DEC, measures are equal ∠BAC ≅ ∠DEC, angles are congruent

Consider a point B on one side of AC. The rays of the form AB can be matched one to one with the real numbers from The measure of  BAC is equal to the absolute value of the difference between the real numbers for AC and AB. C A B

A D C B Applying the Protractor Postulate m∠ CAB = 60˚, m∠ BAD = 130˚ m∠ CAD = ∣ 130˚ - 60˚ ∣ or ∣ 130˚ - 60˚ ∣ = 70˚

A A point is interior of an angle if it is between the sides of the angle A point is exterior of an angle if it is not on the angle or in its interior C B Point C is interior of angle A Point B is exterior of angle A

H J M K If M is in the interior of ∠ JHK, then m∠ JHM + m∠ MHK = m∠ JHK

Draw a sketch using the following information D is interior of ∠ABC C is interior of ∠DBE m ∠ABC = 75˚ m ∠DBC = 45˚ m ∠ABE = 100˚ Find the m ∠ABD, m ∠CBE and m ∠DBE Ans →

D is interior of ∠ABC C is interior of ∠DBE m ∠ABC = 75˚ m ∠DBC = 45˚ m ∠ABE = 100˚ A B C D E 75˚ 45˚ 100˚ Find the m ∠ABD, m ∠CBE and m ∠DBE m∠ABC - m∠DBC = m∠ABD 75˚ - 45˚ = 30˚ m∠ABE - m∠ABC = m∠CBE 100˚ - 75˚ = 25˚ m∠ABD = 30˚ m∠CBE = 25˚ m∠DBC + m∠CBE = m∠DBE 45˚ + 25˚ = 70˚ m∠DBE = 70˚

Angles are classified according to their measures. Angles have measures greater than 0° and less than or equal to 180°

Two angles are adjacent if they share a common vertex and side, but no common interior points. H J M K ∠JHM and ∠MHK are adjacent angles ∠JHM and ∠JHK are not adjacent angles

Bisecting a Segment The midpoint of a segment is the point that bisects the segment, dividing the segment into two congruent segments. Bisect: to divide into two equal parts. A segment bisector is a segment, ray, line or plane that intersects a segment at its midpoint. A B C Congruent segments are indicated using marks through the segments. A B C X Y XY is a bisector of AB If C is the midpoint of AB, then AC ≅ CB

Midpoint Formula To find the coordinates of the midpoint of a segment you find the mean of the x coordinates and the y coordinates of the endpoints. The midpoint of AB = B (10, 6) A ( 3,2 ) A B The midpoint of AB =

Finding the coordinates of an endpoint The midpoint of GH is M ( 7, 5 ). One endpoint is H ( 15, -1 ). Find the coordinates of point G. The x coordinate of G The y coordinate of G G is at ( -1, 11 ) H ( 15, -1 ) M ( 7, 5 ) G ( x₂, y₂ )

Practice Problems JK has endpoints J ( -1, 7 ) and K ( 3, -3 ) find the coordinates of the midpoint NP has midpoint M ( -8, -2 ) and endpoint N ( -5, 9 ). Find the coordinates of P P ( -11, -13 ) ( 1, 2 )

Angle Bisector An angle bisector is a ray that divides an angle into two congruent adjacent angles. C A B D Congruent angles are indicated by separate arcs on each angle If BD is a bisector of ∠ ABC, then ∠ ABD ≅ ∠ DBC

Practice Problems KL is a bisector of ∠ JKM. Find the two angle measures not given in the diagram J M L K 85⁰ J M L K m ∠ JKL = 85/2 = 42½⁰ m ∠ LKM = 85/2 = 42½⁰ 37⁰ m ∠ JKL = m ∠ LKM = 37⁰ m ∠ JKM = 2 x 37 = 74⁰

J M L K KL is a bisector of ∠ JKM. Find the value of x Practice Problem ( 10x – 51 )⁰ ( 6x – 11 )⁰ 6x – 11 = 10x – = 4x = 4x x = 10 Check 6 (10) – 11 = 10 (10) = 100 – = 49 END

Ruler Postulate The points on a line can be matched one to one with real numbers. The real number that corresponds to a point is the coordinate of the point. The distance between points A and B, written as AB, is the absolute value of the difference between the coordinates of A and B. 9 A B 3 Find AB: AB= 3-9 or 9-3 AB=6

Vertical Angles: angles whose sides form opposite rays ∠ 1 and ∠ 3 are vertical angles ∠ 2 and ∠ 4 are vertical angles

Linear Pair: two adjacent angles whose noncommon sides are opposite rays ∠ 1 and ∠ 2 are a linear pair ∠ 2 and ∠ 3 are a linear pair ∠ 3 and ∠ 4 are a linear pair ∠ 4 and ∠ 1 are a linear pair

If m ∠ 1 = 110°, what is the m ∠ 2 and m ∠ 3 and m ∠ 4? What conclusions can you draw about vertical angles? What conclusions can you draw about linear pairs? ˚ 70˚

Practice Problem ( 4x + 15 )° ( 3y + 15 )° ( 3y - 15 )° ( 5x + 30 )° Find the value of x and y ( 4x + 15 ) + ( 5x + 30 ) = 180 9x + 45 = 180 9x = 135 X = 15 ( 3y + 15 ) + ( 3y - 15 ) = 180 6y = 180 Y = 60

Complementary angles: two angles whose sum is 90°. Each of the angles is called the complement of the other. Complementary angles can be adjacent or nonadjacent 23° 67° 23° Adjacent complementary angles Nonadjacent complementary angles

Supplementary angles: two angles whose sum is 180°. Each of the angles is called the supplement of the other. Supplementary angles can be adjacent or nonadjacent 130°50° 130° 50° Adjacent supplementary angles Nonadjacent supplementary angles

Practice Problems T and S are supplementary. The measure of T is half the measure of S. Find the measure of S T and S are complementary. The measure of T is four times the measure of S. Find the measure of S m∠T + m∠S = 180 ½m∠S + m∠S =180 (3/2)∠S =180 m∠S = 120˚ ½m∠S = m∠T m∠T + m∠S = 90 4m∠S = m∠T 4m∠S + m∠S = 90 5m∠S = 90 m∠S = 18˚

Rectangle ( I = length, w = width ) Perimeter: P = 2l + 2w l w Area: A = l w 12 in 5 in P = 2(12) + 2(5) P = P = 34 in A = 12(5) A = 60 in²

Find the perimeter and area of each rectangle 9ft 4ft 15ft A rectangle has a perimeter of 32 in and a length of 9 in. Find its area. 9ft P = 2l + 2w 32 = 2(9) + 2w 32 = w 14 = 2w w = 7 in P = 2(9) + 2(4) P = P = 26 ft A = 12(9) A = 108 ft² A = 9(7) A = 63 in² Practice Problems A = 9(4) A = 36 ft² P = 2(12) + 2(9) P = P = 42 ft

Square (s = side length) Perimeter: P = 4s Area: s² Find the perimeter and area of each square 8ft 4m Find the perimeter of a square with an area of 64 ft²

Find the perimeter and area of each square 8ft 4m Find the perimeter of a square with an area of 81 ft² a = b a² + a² = c² 2a² = 4² 2a² = 16 a² = 8 a = √ 8 s² = 81 s = 9 P = 4(8) P = 32 ft A = 8² A = 64 ft² P = 4√ 8 P = 4(2)√ 2 P = 8√2 m A = (√8)² A = 8m² P = 4(9) P = 36 ft²

Triangle ( a, b, c = side lengths, b = base, h = height) Perimeter: P = a + b + c Area: A = ½ bh Find the perimeter and area of each triangle: 4m 5m 15m 14m 7 ft

Find the perimeter and area of each triangle: 4m 5m 15m 14m 7 ft ft P = P = 34 m A = ½ 15(4) A = 30 m² P = P = 14 + A = ½ 7 (7) A = 24½ ft²

Circle ( r = radius) Find the circumference and area of each circle 6in 8 ft Circumference: C = 2 π r Area: A = πr²

R = 8/2 = 4 C = 2 π 4 C = 8 π C ≈ ft A = π 4² A = 16 A ≈ ft² Find the circumference and area of each circle 6in 8 ft C = 2 π 6 C = 12 π C ≈ in A = π 6² A = π 36 A ≈ in²

Problem Solving Steps 1.Define the variable(s) 2.Write an equation using the variable 3.Solve the equation 4.Answer the question

Practice Problem A painter is painting one side of a wooden fence along a highway. The fence is 926 ft long and 12 ft tall. Each five gallon can of paint can cover 2000 square feet. How many cans of paint will be needed to paint the fence. 1.Define the variable: x = number of 5 gallon cans needed 2. Write an equation using the variable: 2000x = 926(12) 3. Solve the equation 2000x = 926(12) 2000x = x = Answer the question: 6 cans of paint will be needed to paint the fence