geometry 3rd-4th grading lessons

Summary of the

Properties of Special

Quadrilaterals

In a RECTANGLE

1. Opposite sides are congruent

2. Opposite sides are parallel3. Each diagonal separates the

rectangle into two congruent triangles

4. Opposite angles are congruent

5. Consecutive angles are supplementary

6. All angles are right angles.7. Diagonals bisect each other

and are congruent.

In a RHOMBUS

1. All the sides are congruent2. Opposite sides are parallel3. Each diagonals bisect the

rhombus into two congruent triangles

4. Opposite angles are congruent

5. Consecutive angles are congruent

6. Diagonals bisect each other and are perpendicular

In a SQUARE

1. All the sides are congruent2. All angles are right angles3. Each diagonals separate the

square into two congruent triangles

4. Opposite angles are congruent and are supplementary

5. Consecutive angles are congruent

6. Diagonals bisect each other and are perpendicular and are congruent

PROPERTIES OF PARALLELOGRAM

1.Opposite angles are congruent

2. Two non-opposite angles of a parallelogram are supplementary

3. Opposite sides are congruent

4. A diagonal of a parallel divides the parallelogram into congruent angles

5. Diagonals bisect each other

Corresponding Parts of

Congruent Triangles are Congruent(CPCTC)

Conditions for the

Congruence Between Two

Right Triangles

LL = Legs Legs

HyL = Hypotenuse Legs

HyA = Hypotenuse Acute

LA = Legs Acute

Ratio and Proportion

s

RATIO

• Comparison of two quantities.

WAYS OF WRITING

• BY COLON5 : 6 o r a : b

• BY FRACTION5 / 6 o r a / b

• BY DIVISION5 ÷ 6 or a ÷ b

PROPORTION

• An equation in which 2 quantities are set equal

a:b = c:d

abcd

EXTREMES

MEANS

Proportional Segments

PROPORTIONAL SEGMENTS

&BASIC SIMILARITY

THEOREM

ILLUSTRATION

Pt. X divides segment AB so that AX to XB is 3 : 2.

Pt. Y divides segment CD so that CY to YD is 3 : 2

A

DYC

X B12 8

6 4

THEOREM: PROPORTIONAL SEGMENTS “Two segments are divided

proportionally if the measures of the segments of one have the same ratio as the measures of the corresponding segments of the other.”

A

DYC

X B12 8

6 4AX CY

12

XB YD

6 8 4

SOME PROPORTIONS

3 3

2 2

1.



A

DYC

X B12 8

6 4AB CD

20

XB YD

108 4

SOME PROPORTIONS

5 5

2 2

2.



A

DYC

X B12 8

6 4AB CD

20

AX CY

1012 6

SOME PROPORTIONS

5 5

3 3

3.



A

DYC

X B12 8

6 4AX +AB CY +CD

32

AX CY

1612 6

SOME PROPORTIONS

8 8

3 3

4.

Illustrative Examples

Suppose segment AC and segment MP are divided proportionally by points B and N respectively. Then,

A

PNM

B C8 12

2 3AB MN

AB

BC NP

BCMN NP

1.

2.

Suppose segment AC and segment MP are divided proportionally by points B and N respectively. Then,

A

PNM

B C8 12

2 3AB MN

BC

AC MP

NPAC MP

3.

4.

Find the unknown parts assuming the segments are divided proportionally.

X 32

6 8

Solution:

X : 32 = 6 : 8

Applying the law of proportion

8(x) = 6( 32)

8x = 192

X = 24

Solution: Let x = the length of the other

portion of the second string.

x 60

15 18

x18

60

6015

5x6

6x 5(60)

X = 50, the length of the other portion of the second string

6x = 300

BASIC PROPORTIONALITY THEOREM

If a line intersects two sides of a triangle and is parallel to the third side, then it divides the first two sides proportionally.

RESTATEMENT OF THE THEOREM

If a line (EF) intersects two sides ( AB & CB) of a triangle (ABC) and is parallel to the third side( AC ), then it divides the first two sides proportionally.

Thus,

B

E F

CA

OTHER PROPORTIONS

1. BE : EA = BF : FC

2. BE : BA = BF : BC

3. BA : EA = BC : FC

4. BE : BF = EA : FC

5. FC : EA = BC : BA

6. EF : AC = BF : BC

7. EF : AC = BE : BA

B

E F

CA

VERIFYING A PROPORTIONS( an example)

1. BE : EA = BF : FC

15 : 5 = 12 : 4 By simplifying, 3 : 1 = 3 : 1

B

E F

CA

15

4

12

58

6

VERIFYING A PROPORTIONS

2. BE : BA = BF : BC

15 : 20 = 12 : 16

By simplifying,3 : 4 = 3 : 4

B

E F

CA

15

4

12

58

6


3. BA : EA = BC : FC

20 : 5 = 16 : 4

By simplifying,4 : 1 = 4 : 1

B

E F

CA

15

4

12

58

6


4. BE : BF = EA : FC

15 : 12 = 5 : 4 By simplifying,

5 : 4 = 5 : 4

B

E F

CA

15

4

12

58

6


5. FC : EA = BC : BA


4 : 5 = 4 : 5

B

E F

CA

15

4

12

58

6


6. EF : AC = BF : BC


3 : 4 = 3 : 4

B

E F

CA

15

4

12

58

6


6. EF : AC = BE : BA


3 : 4 = 3 : 4

B

E F

CA

15

4

12

58

6

Exercises

GIVEN: DE // BC, AD = 9, AE = 12,

DE = 10,DB = 18.Find,BC, AC and CE.

A

D E

CB

9 12

18

10

Solution Find BC,BC : DE = BA : DABC : 10 = 27 : 9 orBC : 10 = 3 : 1Applying principle

of proportionBC(1) = 10(3)BC = 30

A

D E

CB

9 12

18

10

30

Solution Find AC,AC : AE = BA : DAAC : 12 = 27 : 9 orAC : 12 = 3 : 1Applying principle

of proportionAC(1) = 12(3)AC = 36

A

D E

CB

9 12

18

10

Solution Find CE,CE : AE = BD : DACE : 12 = 18 : 9 orCE : 12 = 2 : 1Applying principle

of proportionCE(1) = 12(2)CE = 24

A

D E

CB

9 12

18

10

24

Solution Another way to

find CE,CE = AC - AEHence, AC =36,

thenCE = 36 - 12CE = 24

A

D E

CB

9 12

18

10

24

Converse of Basic

Proportionality Theorem

• If a line divides any two sides of a triangle in the same ratio, then the linemust be parallel to the third side.

Segment cut by a transversal

corollary

Definition

In mathematics, a corollary is a statement which follows readily from a previously proven statement, typically a mathematical theorem.

http://en.wikipedia.org/wiki/Mathematics

http://en.wikipedia.org/wiki/Theorem

Segment cut by a transversal

If three parallel lines intersect two transversals, then they divide the transversals proportionally.

Restatement of the corollary

If AD //EF//BC and intersect two transversals

(line AB and line DC) , then

DC : DF = AB : AE BE : EA = CF : FD AE : DF = BE : CF

B

DA

C

If three parallel lines intersect two transversals, then they divide the

transversals proportionally

FE

Find the value of x. Solution:By applying the corollary 10 : x = 12 : x + 2Using the principle of

proportion 10(x +2) = x ( 12 ) 10x + 20 = 12x 20 = 12x – 10x 20 = 2x X = 10

X + 2

10 12

x

Illustrative examples

By applying the theorem

x : x + 2 = 10 : 12 orx : x + 2 = 5 : 6Using the principle of

proportion 6(x) = 5 (x + 2 ) 6x = 5x + 10 6x – 5x = = 10 X = 10

X + 2

10 12

x

Another solution

Segment cut by angle bisector

Theorem

Segment cut by angle bisector

If a ray bisects an angle of a triangle, it divides the opposite side into segments proportional to the other two sides.

Exploration Construct any triangle. Construct

an angle bisector in the triangle and draw the segment along the angle bisector from the vertex to the intersection with the opposite side.

Measure the ratio of the adjacent sides .

Measure the ratio of the segments cut off by the bisector on the opposite side.

Repeat for many triangles .

ILLUSTRATION

The bisector of an angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides.

ILLUSTRATION

for any triangle ABC, the bisector of the angle at C divides the opposite side into segments of length x and y such that A D B

C

ILLUSTRATION

or

x : y = a : b

A D B

C

xy b

a

Restatement of the theorem

If AD bisects angle BAC of triangle ABC, then …

BD : DC = AB : AC BD : BC = AB : AB +

AC DC : BC = CA : CA +

ABB D

A

C

If a ray bisects an angle of a triangle, it divides the opposite side into segments proportional to the other two sides.

Find the value of x if

a =10, b = 15 and

y = 12.

Illustrative examples


x : y = a : b or

x : 12 = 10 : 15Using the principle of

proportion 15(x) = 12 (10 ) 15x = 120 x = 120 15 x = 8

solution


x : a = y : b or

x : 10 = 12 : 15Using the principle of

proportion 15(x) = 12 (10 ) 15x = 120 x = 120 15 x = 8

another solution


x : x + y = a : a +b or

x : x+12= 10 : 10 +15

x : x+12= 10 : 25

x : x+12= 2: 5Using the principle of

proportion 5(x) = 2 (x +12 ) 5x = 2x + 24 5x – 2x = 24 3x = 24 x = 8

another solution

Similar Triangles

SIMILAR TRIANGLES

Two triangles are similar if their corresponding angles are congruent and the measures of corresponding triangles are proportional.

~ MEANS SIMILAR

PythagoreanTheorem

PYTHAGOREAN THEOREM

The PYTHAGOREAN THEOREM is used to find the length of a missing side in a right triangle.

45-45-90And

30-60-90TRIANGLES

45-45-90

• A 45°- 45°- 90° triangle is a special right triangle whose angles are 45°, 45°and 90°. The lengths of the sides of a 45°- 45°- 90° triangle are in the ratio of . Note that a 45°- 45°- 90° triangle is an isosceles right triangle.

30-60-90• Another type of special right

triangles is the 30°- 60°- 90° triangle. This is right triangle whose angles are 30°, 60°and 90°. The lengths of the sides of a 30°- 60°- 90° triangle are in the ratio of

• The hypotenuse is always twice the length of the shorter leg (the side facing the 30° angle). The longer leg (the side facing the 60° angle) is times of the shorter leg.

Central Angle, Inscribed Angle and

Intercepted Arc

CENTRAL ANGLE

• A central angle is an angle formed by two intersecting radii such that its vertex is at the center of the circle.

<AOB is a central angle.

Its intercepted arc is the minor arc from A to B.

m<AOB = 80°

INSCRIBED ANGLE

• An inscribed angle is an angle with its vertex "on" the circle, formed by two intersecting chords.

<ABC is an inscribed angle.

Its intercepted arc is the minor arc from A to C.

m<ABC = 50°

INTERCEPTED ARC

• Corresponding to an angle, this is the portion of the circle that lies in the interior of the angle together with the endpoints of the arc.

∠ ABC is an inscribed angle and is its intercepted arc.

THEOREMS• The measure of an inscribed angle

in a circle equals half the measure of its intercepted arc.

• If two inscribed angles of a circle intercept the same arc or arcs of equal measure, then the inscribed angles have equal measure.

• If an inscribed angle intercepts a semicircle, then its measure is 90°.

Cartesian Coordinate System

Origin X – AxisY – Axis

Quadrants

Ordered Pair

Coordinates

Terminologies:

Cartesian Coordinate plane

a two-dimensional surface on which a coordinate system has been set up; invented by Rene Descartes; also called Cartesian plane, graph, coordinate grid.

Coordinates

the numbers in an ordered pair that locate a point in the coordinate plane.

Ordered pair

a pair of numbers, written as

(x, y), that represents a point on a coordinate grid.

X-axis

the horizontal number line on a coordinate grid.

Y-axis

the vertical number line on a coordinate grid

Quadrants

The four parts made by X and Y axis

Origin

Intersection of X and Y axis

-x x

-y

y

Origin

1 2 3

-1

-2

-3

1

2

3

-1-2-3

Quadrant 1Quadrant 2

Quadrant 4Quadrant 3

( + , + )( - , + )

( - , - ) ( + , - )

-x x

-y

y

1 2 3

-1

-2

-3

1

2

3

-1-2-3

Plot the point:

1. ( 1, 2)

-x x

-y

y

1 2 3

-1

-2

-3

1

2

3

-1-2-3

Plot the point:

2. ( 2, -3)

-x x

-y

y

1 2 3

-1

-2

-3

1

2

3

-1-2-3

Plot the point:

2. ( -2, 3)

-x x

-y

y

1 2 3

-1

-2

-3

1

2

3

-1-2-3

Plot the point:

2. (-2,-3)

LINEAR EQUATIONS PART I

1. Basic Coordinate Plane Info2. Review on Plotting Points

3. Finding Slopes4. x and y intercepts

5. Slope-Intercept Form of a Line6. Graphing Lines

7.Determine the equation of a line given two points, slope and one point, or a graph.

Assignments

COORDINATE PLANE

Parts of a plane1. X-axis2. Y-axis3. Origin4. Quadrants I-IV

X-axis

Y-axis

Origin ( 0 , 0 )

QUAD IQUAD II

QUAD III QUAD IV

PLOTTING POINTSRemember when plotting points you always start at the origin. Next you go left (if x-coordinate is negative) or right (if x-coordinate is positive. Then you go up (if y-coordinate is positive) or down (if y-coordinate is negative)

Plot these 4 pointsA (3, -4), B (5, 6), C (-4, 5) and D (-7, -5)

A

BC

D

SLOPESlope is the ratio of the vertical rise to the horizontal run between any two points on a line. Usually referred to as the rise over run. Slope triangle between two

points. Notice that the slope triangle can be drawn two different ways.

Rise is -10 because we went down

Run is -6 because we went to the left

3

5

6

10

iscasethisinslopeThe

Rise is 10 because we went up

Run is 6 because we went to the right

3

5

6

10iscasethisinslopeThe

Another way to find slope

FORMULA FOR FINDING SLOPE

21

21

12

12

YY

XX

YY

XX

RUN

RISESLOPE

The formula is used when you know two points of a line.

),(),( 2211 YXBandYXAlikelookThey

EXAMPLE

Find the slope of the line between the two points (-4, 8) and (10, -4)

If it helps label the points. 1X 1Y2X 2Y

Then use the formula

12

12

YY

XX

)8()4(

)4()10(

FORMULAINTOSUBSTITUTE

6

7

12

14

)8(4

410

)8()4(

)4()10(

SimplifyThen

X AND Y INTERCEPTSThe x-intercept is the x-coordinate of a point where the graph crosses the x-axis.

The y-intercept is the y-coordinate of a point where the graph crosses the y-axis.

The x-intercept would be 4 and is located at the point (4, 0).

The y-intercept is 3 and is located at the point (0, 3).

SLOPE-INTERCEPT FORM OF A LINEThe slope intercept form of a line is y = mx + b, where “m” represents the slope of the line and “b” represents the y-intercept.

When an equation is in slope-intercept form the “y” is always on one side by itself. It can not be more than one y either.

If a line is not in slope-intercept form, then we must solve for “y” to get it there.

Examples

IN SLOPE-INTERCEPT NOT IN SLOPE-INTERCEPT

y = 3x – 5 y – x = 10

y = -2x + 10 2y – 8 = 6x

y = -.5x – 2 y + 4 = 2x

Put y – x = 10 into slope-intercept form

Add x to both sides and would get y = x + 10

Put 2y – 8 = 6x into slope-intercept form.

Add 8 to both sides then divide by 2 and would get y = 3x + 4

Put y + 4 = 2x into slope-intercept form.

Subtract 4 from both sides and would get y = 2x – 4.

GRAPHING LINESBY MAKING A TABLE OR USING THE

SLOPE-INTERCEPT FORM

I could refer to the table method by input-output table or x-y table. For now I want you to include three values in your table. A negative number, zero, and a positive number.

Graph y = 3x + 2 INPUT (X) OUTPUT (Y)

-2 -4

0 2

1 5

By making a table it gives me three points, in this case (-2, -4) (0, 2) and (1, 5) to plot and draw the line.

See the graph.

Plot (-2, -4), (0, 2) and (1, 5)

Then draw the line. Make sure your line covers the graph and has arrows on both ends. Be sure to use a ruler.

Slope-intercept graphing

Slope-intercept graphingSteps1. Make sure the equation is in slope-intercept form.2. Identify the slope and y-intercept.3. Plot the y-intercept.4. From the y-intercept use the slope to get another point to draw the line.

1. y = 3x + 22. Slope = 3 (note that this means the

fraction or rise over run could be (3/1) or (-3/-1). The y-intercept is 2.

3. Plot (0, 2)4. From the y-intercept, we are going rise

3 and run 1 since the slope was 3/1.

FIND EQUATION OF A LINE GIVEN 2 POINTS

1. Find the slope between the two points.

2. Plug in the slope in the slope-intercept form.

3. Pick one of the given points and plug in numbers for x and y.

4. Solve and find b.5. Rewrite final form.

Find the equation of the line between (2, 5) and (-2, -3).

1. Slope is 2.2. y = 2x + b3. Picked (2, 5) so

(5) = 2(2) + b4. b = 15. y = 2x + 1

Two other ways

Steps if given the slope and a point on the line.1. Substitute the slope into

the slope-intercept form.

2. Use the point to plug in for x and y.

3. Find b.4. Rewrite equation.

If given a graph there are three ways.

One way is to find two points on the line and use the first method we talked about.

Another would be to find the slope and pick a point and use the second method.

The third method would be to find the slope and y-intercept and plug it directly into y = mx + b.

Solving Systems Of

Linear Equations

Systems of Linear Equations

Using a Graph to Solve

All the slides in this presentation are timed.

You do not need to click the mouse or press any keys on the keyboard for the presentation on each slide to continue.

However, in order to make sure the presentation does not go too quickly, you will need to click the mouse or press a key on the keyboard to advance to the next slide.

You will know when the slide is finished when you see a small icon in the bottom left corner of the slide.

Click the mouse button to advance the slide when you see this icon.

What is a System of Linear Equations?A system of linear equations is simply two or more linear equations using the same variables.

We will only be dealing with systems of two equations using two variables, x and y.

If the system of linear equations is going to have a solution, then the solution will be an ordered pair (x , y) where x and y make both equations true at the same time.

We will be working with the graphs of linear systems and how to find their solutions graphically.

If the lines are parallel, there will be no solutions.If the lines are the same, there will be an infinite number of solutions.

How to Use Graphs to Solve Linear Systems

x

yConsider the following system:

x – y = –1

x + 2y = 5

Using the graph to the right, we can see that any of these ordered pairs will make the first equation true since they lie on the line.

We can also see that any of these points will make the second equation true.

However, there is ONE coordinate that makes both true at the same time…

(1 , 2)

The point where they intersect makes both equations true at the same time.

• If the lines cross once, there will be one solution.

• If the lines are parallel, there will be no solutions.

• If the lines are the same, there will be an infinite number of solutions.

x – y = –1

x + 2y = 5

How to Use Graphs to Solve Linear Systems

x

yConsider the following system:

(1 , 2)

We must ALWAYS verify that your coordinates actually satisfy both equations.

To do this, we substitute the coordinate (1 , 2) into both equations.

x – y = –1

(1) – (2) = –1 Since (1 , 2) makes both equations true, then (1 , 2) is the solution to the system of linear equations.

x + 2y = 5

(1) + 2(2) =

1 + 4 = 5

Graphing to Solve a Linear System

While there are many different ways to graph these equations, we will be using the slope - intercept form.

To put the equations in slope intercept form, we must solve both equations for y.

Start with 3x + 6y = 15

Subtracting 3x from both sides yields

6y = –3x + 15

Dividing everything by 6 gives us…

512 2y x=- +

Similarly, we can add 2x to both sides and then divide everything by 3 in the second equation to get

23 1y x= -

Now, we must graph these two equations.

Solve the following system by graphing:

3x + 6y = 15

–2x + 3y = –3


512 2

23 1

y x

y x

=- +

= -

Solve the following system by graphing:

3x + 6y = 15

–2x + 3y = –3

Using the slope intercept form of these equations, we can graph them carefully on graph paper.

x

y

Start at the y - intercept, then use the slope.Label the solution!

(3 , 1)

Lastly, we need to verify our solution is correct, by substituting (3 , 1).

Since and , then our solution is correct!( ) ( )3 3 6 1 15+ = ( ) ( )2 3 3 1 3- + =-


Let's summarize! There are 4 steps to solving a linear system using a graph.

Step 1: Put both equations in slope - intercept form.

Step 2: Graph both equations on the same coordinate plane.

Step 3: Estimate where the graphs intersect.

Step 4: Check to make sure your solution makes both equations true.

Solve both equations for y, so that each equation looks like

y = mx + b.

Use the slope and y - intercept for each equation in step 1. Be sure to use a ruler and graph paper!

This is the solution! LABEL the solution!

Substitute the x and y values into both equations to verify the point is a solution to both equations.

x

y

LABEL the solution!


Step 1: Put both equations in slope - intercept form.

Step 2: Graph both equations on the same coordinate plane.

Step 3: Estimate where the graphs intersect. LABEL the solution!

Step 4: Check to make sure your solution makes both equations true.

Let's do ONE more…Solve the following system of equations by graphing.

2x + 2y = 3

x – 4y = -1

32y x=- +

1 14 4y x= +

( ) ( )122 1 2 2 1 3+ = + =

( )121 4 1 2 1- = - =-

( )121,

3-2: Solving Systems of Equations using Substitution

Solving Systems of Equations using Substitution

Steps:

1. Solve one equation for one variable (y= ; x= ; a=)

2. Substitute the expression from step one into the other equation.

3. Simplify and solve the equation.

4. Substitute back into either original equation to find

the value of the other variable.

5. Check the solution in both equations of the system.

Example #1: y = 4x3x + y = -21

Step 1: Solve one equation for one variable.

y = 4x (This equation is already solved for y.)

Step 2: Substitute the expression from step one into the other equation.

3x + y = -21

3x + 4x = -21

Step 3: Simplify and solve the equation.

7x = -21

x = -3

y = 4x3x + y = -21

Step 4: Substitute back into either original equation to find the value of the other variable.

3x + y = -21 3(-3) + y = -21 -9 + y = -21 y = -12

Solution to the system is (-3, -12).

y = 4x3x + y = -21

Step 5: Check the solution in both equations.

y = 4x

-12 = 4(-3)

-12 = -12

3x + y = -21

3(-3) + (-12) = -21

-9 + (-12) = -21

-21= -21

Solution to the system is (-3,-12).

Example #2: x + y = 10 5x – y = 2

Step 1: Solve one equation for one variable.

x + y = 10

y = -x +10Step 2: Substitute the expression from step one into

the other equation.

5x - y = 2

5x -(-x +10) = 2

x + y = 10 5x – y = 2

5x -(-x + 10) = 2

5x + x -10 = 2

6x -10 = 2

6x = 12

x = 2

Step 3: Simplify and solve the equation.

x + y = 10 5x – y = 2

Step 4: Substitute back into either original equation to find the value of the other variable.

x + y = 102 + y = 10 y = 8

Solution to the system is (2,8).

x + y = 10 5x – y = 2

Step 5: Check the solution in both equations.

x + y =10

2 + 8 =10

10 =10

5x – y = 2

5(2) - (8) = 2

10 – 8 = 2

2 = 2

Solution to the system is (2, 8).

Solving Systems of Equations

The Elimination Method

Objectives

• Learn the procedure of the Elimination Method using addition

• Learn the procedure of the Elimination Method using multiplication

• Solving systems of equations using the Elimination Method

Elimination using Addition

Consider the system

x - 2y = 5

2x + 2y = 7

REMEMBER: We are trying to find the Point of Intersection. (x, y)

Lets add both equations to each other


Consider the system

x - 2y = 5

2x + 2y = 7

Lets add both equations to each other+

NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.


Consider the system

x - 2y = 5

2x + 2y = 7

Lets add both equations to each other+

3x = 12x = 4

ANS: (4, y)



Consider the system

x - 2y = 5

2x + 2y = 7

ANS: (4, y)

Lets substitute x = 4 into this equation.

4 - 2y = 5 Solve for y - 2y = 1

y = 12



Consider the system

x - 2y = 5

2x + 2y = 7

ANS: (4, )

Lets substitute x = 4 into this equation.

4 - 2y = 5 Solve for y - 2y = 1

y = 12

12



Consider the system

3x + y = 14

4x - y = 7



Consider the system

3x + y = 14

4x - y = 7

7x = 21x = 3

ANS: (3, y)

+


Consider the system

ANS: (3, )

3x + y = 14

4x - y = 7

Substitute x = 3 into this equation

3(3) + y = 149 + y = 14

y = 5

5


Examples…

2x y+ 5=

3x y 15=

1. 2.

2y x 5=

6y x+ 11=

ANS: (4, -3) ANS: (-1, 2)

Elimination using Multiplication

Consider the system

6x + 11y = -5

6x + 9y = -3


Consider the system

6x + 11y = -5

6x + 9y = -3+12x + 20y = -8 When we add equations together,

nothing cancels out


Consider the system

6x + 11y = -5

6x + 9y = -3


Consider the system

6x + 11y = -5

6x + 9y = -3

-1 ( )


Consider the system

- 6x - 11y = 5

6x + 9y = -3+-2y = 2

y = -1

ANS: (x, )-1


Consider the system

6x + 11y = -5

6x + 9y = -3

ANS: (x, )-1

y = -1

Lets substitute y = -1 into this equation

6x + 9(-1) = -36x + -9 = -3

+9 +9

6x = 6x = 1


Consider the system

6x + 11y = -5

6x + 9y = -3

ANS: ( , )-1

y = -1

Lets substitute y = -1 into this equation

6x + 9(-1) = -36x + -9 = -3

+9 +9

6x = 6x = 1

1


Consider the system

x + 2y = 6

3x + 3y = -6

Multiply by -3 to eliminate the x term


Consider the system

x + 2y = 6

3x + 3y = -6

-3 ( )


Consider the system

-3x + -6y = -18

3x + 3y = -6+-3y = -24

y = 8

ANS: (x, 8)


Consider the system

x + 2y = 6

3x + 3y = -6

ANS: (x, 8)

Substitute y =14 into equation

y =8

x + 2(8) = 6x + 16 = 6

x = -10


Consider the system

x + 2y = 6

3x + 3y = -6

ANS: ( , 8)

Substitute y =14 into equation

y =8

x + 2(8) = 6x + 16 = 6

x = -10

-10

Examples

1.x + 2y = 5

2x + 6y = 12

2.

ANS: (3, 1)

x + 2y = 4

x - 4y = 16

ANS: (8, -2)

More complex ProblemsConsider the system

3x + 4y = -25

2x - 3y = 6

Multiply by 2

Multiply by -3


3x + 4y = -25

2x - 3y = 6

2( )

-3( )


6x + 8y = -50

-6x + 9y = -18+17y = -68

y = -4

ANS: (x, -4)


3x + 4y = -25

2x - 3y = 6

ANS: (x, -4)

Substitute y = -4

2x - 3(-4) = 62x - -12 = 6

2x + 12 = 6

2x = -6

x = -3


3x + 4y = -25

2x - 3y = 6

ANS: ( , -4)

Substitute y = -4

2x - 3(-4) = 62x - -12 = 6

2x + 12 = 6

2x = -6

x = -3 -3

Examples…

1. 2.

4x + y = 9

3x + 2y = 8

2x + 3y = 1

5x + 7y = 3

ANS: (2, 1) ANS: (2, -1)

Parallel and Perpendicular Lines

Parallel Lines

Two lines with the same slope are said to be parallel lines. If you graph them they will never intersect.

We can decide algebraically if two lines are parallel by finding the slope of each line and seeing if the slopes are equal to each other.

We can find the equation of a line parallel to a given line and going through a given point by:

a.) first finding the slope m of the given line; b.) finding the equation of the line through the given point with slope m.

Testing if Lines are Parallel

Are the lines parallel?

12 3 9 and -8 2 14x y x y

Find the slope of 12 3 9

3 12 9

4 3

x y

y x

y x

The slope m = -4


2 8 14

4 7

x y

y x

y x

The slope m = -4

Since the slopes are equal the lines are parallel.

Graphs of Parallel Lines

The red line is the graph of y = – 4x – 3 and the blue line is the graph ofy = – 4x – 7

Practice Testing if Lines are Parallel

Are the lines 6 3 5 and 2 4 4x y y x parallel? (click mouse for answer)

6 3 5

3 6 5

52 32

x y

y x

y x

m

2 4 4

2 2

2

y x

y x

m

Since the slopes are differentthe lines are not parallel.

Are the lines 2 4 and 2 4 12x y x y parallel? (click mouse for answer)

2 4

2 4

1 221

2

x y

y x

y x

m

2 4 12

4 2 12

1 321

2

x y

y x

y x

m

Since the slopes are equalthe lines are parallel.

Constructing Parallel Lines

Find the equation of a line going through the point (3, -5) and parallel to 2 83y x

Using the point-slope equation where the slope m = -2/3 and

the point is (3, -5) we get 25 33

25 232 33

y x

y x

y x

Practice Constructing Parallel Lines

Find the equation of the line going through the point (4,1) and parallel to (click mouse for answer) 3 7y x

1 3 4

1 3 12

3 13

y x

y x

y x

Find the equation of the line going through the point (-2,7) and parallel to (click mouse for answer) 2 8x y

7 2 2

7 2 2

7 2 4

2 3

y x

y x

y x

y x

Perpendicular Lines

Perpendicular lines are lines that intersect in a right angle. We can decide algebraically if two lines are perpendicular by finding

the slope of each line and seeing if the slopes are negative reciprocals of each other. This is equivalent to multiplying the two slopes together and seeing if their product is –1.

We can find the equation of a line perpendicular to a given line and going through a given point by:

a.) first finding the slope m of the given line;

b.) finding the equation of the line through the given point

with slope = –1 /m.

Testing if Lines Are Perpendicular

1Are the lines 2 5 and 4 perpendicular?

2x y y x


2 5

x y m

y x

1 1Find the slope of 4

2 2y x m

Since the slopes are negative reciprocals of each other the lines are perpendicular. 1

2 12

Graphs of Perpendicular Lines

The red line is the graph of y = – 2x + 5 and the blue line is the graph ofy = – 1/2 x +4

Practice Testing if Lines Are Perpendicular

Are the lines 6 3 5 and 2 4 4 perpendicular?x y y x 6 3 5

3 6 5

52 32

x y

y x

y x

m

2 4 4

2 2

2

y x

y x

m

Since the slopes are not negative reciprocals of each other (their product is not -1) the lines are not perpendicular

Are the lines 2 4 and 4 2 6 perpendicular?x y x y 2 4

2 4

1 221

2

x y

y x

y x

m

4 2 6

2 4 6

2 3

2

x y

y x

y x

m

Since the slopes are negative reciprocals of each other (their product is -1) the lines are perpendicular.

Constructing Perpendicular Lines

Find the equation of a line going through the point (3, -5) and perpendicular to 2 83y x

The slope of the perpendicular line will be m = 3/2 Using

the point-slope equation where the slope m = 3/2 and

the point is (3, -5) we get 35 323 95 2 2

3 192 2

y x

y x

y x

Practice Constructing Perpendicular Lines

Find the equation of the line going through the point (4,1) and perpendicular to (click mouse for answer) 3 7y x

11 431 41 3 31 1

3 3

y x

y x

y x

Find the equation of the line going through the point (-2,7) and perpendicular to (click mouse for answer) 2 8x y

17 2217 2217 121 82

y x

y x

y x

y x

The Distance and Midpoint Formulas

Topic 7.1

Distance Formula

Used to find the distance between two points

2 22 1 2 1distance ( ) ( )x x y y

Example Find the distance between A(4,8) and B(1,12)

2 22 1 2 1distance ( ) ( )x x y y

A (4, 8) B (1, 12)

2 2distance (1 4) (12 8) 2 2distance ( 3) (4)

distance 9 16 25 5

YOU TRY!!

Find the distance between: A. (2, 7) and (11, 9)

B. (-5, 8) and (2, - 4)

2 2(9) (2) 85

2 2(7) ( 12) 193

Midpoint Formula

Used to find the center of a line segment

2 1 2 1midpoint ,2 2

x x y y

Example Find the midpoint between A(4,8) and B(1,12)

2 1 2 1midpoint ,2 2

x x y y

A (4, 8) B (1, 12)

1 4 12 8midpoint ,

2 2

midpoint 5,10

2

YOU TRY!!

Find the midpoint between: A) (2, 7) and (14, 9)

B) (-5, 8) and (2, - 4)

midpoint = 8,8

-3midpoint = , 2

2

Holt Geometry

4-7 Introduction to Coordinate Proof4-7 Introduction to Coordinate Proof

Holt Geometry

Warm Up

Lesson Presentation

Lesson Quiz

Holt Geometry

4-7 Introduction to Coordinate Proof

Warm UpEvaluate.1. Find the midpoint between (0, 2x) and (2y, 2z).

2. One leg of a right triangle has length 12, and the hypotenuse has length 13. What is the length of the other leg?

3. Find the distance between (0, a) and (0, b), where b > a.

(y, x + z)

5

b – a

Holt Geometry


Position figures in the coordinate plane for use in coordinate proofs.

Prove geometric concepts by usingcoordinate proof.

Objectives

Holt Geometry


coordinate proof

Vocabulary

Holt Geometry


You have used coordinate geometry to find the midpoint of a line segment and to find the distance between two points. Coordinate geometry can also be used to prove conjectures.

A coordinate proof is a style of proof that uses coordinate geometry and algebra. The first step of a coordinate proof is to position the given figure in the plane. You can use any position, but some strategies can make the steps of the proof simpler.

Holt Geometry


Holt Geometry


Position a square with a side length of 6 units in the coordinate plane.

Example 1: Positioning a Figure in the Coordinate Plane

You can put one corner of the square at the origin.

Holt Geometry


Check It Out! Example 1

Position a right triangle with leg lengths of 2 and 4 units in the coordinate plane. (Hint: Use the origin as the vertex of the right angle.)

Holt Geometry


Once the figure is placed in the coordinate plane, you can use slope, the coordinates of the vertices, the Distance Formula, or the Midpoint Formula to prove statements about the figure.

Holt Geometry


Example 2: Writing a Proof Using Coordinate Geometry

Write a coordinate proof.

Given: Rectangle ABCD with A(0, 0), B(4, 0), C(4, 10), and D(0, 10)

Prove: The diagonals bisect each other.

Holt Geometry


The midpoints coincide,

therefore the diagonals

bisect each other.

By the Midpoint Formula,

mdpt. of

mdpt. of

AC0 4 0 10

, (2,5)2 2

Example 2 Continued

Holt Geometry



Proof: ∆ABC is a right triangle with height AB and base BC.

Use the information in Example 2 (p. 268) to write a coordinate proof showing that the area of ∆ADB is one half the area of ∆ABC.

area of ∆ABC = bh

= (4)(6) = 12 square units

1

21

2

Holt Geometry


By the Midpoint Formula, the coordinates of

D = = (2, 3). 0+4 , 6+0 2 2

The x-coordinate of D is the height of ∆ADB, and the base is 6 units.

Check It Out! Example 2 Continued

The area of ∆ADB = bh

= (6)(2) = 6 square units

Since 6 = (12), the area of ∆ADB is one half the

area of ∆ABC.

12

12

12

Holt Geometry


A coordinate proof can also be used to prove that a certain relationship is always true.

You can prove that a statement is true for all right triangles without knowing the side lengths.

To do this, assign variables as the coordinates of the vertices.

Holt Geometry


Position each figure in the coordinate plane and give the coordinates of each vertex.

Example 3A: Assigning Coordinates to Vertices

rectangle with width m and length twice the width

Holt Geometry


Example 3B: Assigning Coordinates to Vertices

Position each figure in the coordinate plane and give the coordinates of each vertex.

right triangle with legs of lengths s and t

Holt Geometry


Do not use both axes when positioning a figure unless you know the figure has a right angle.

Caution!

Holt Geometry



Position a square with side length 4p in the coordinate plane and give the coordinates of each vertex.

Holt Geometry


If a coordinate proof requires calculations with fractions, choose coordinates that make the calculations simpler.

For example, use multiples of 2 when you are to find coordinates of a midpoint. Once you have assigned the coordinates of the vertices, the procedure for the proof is the same, except that your calculations will involve variables.

Holt Geometry


Because the x- and y-axes intersect at right angles, they can be used to form the sides of a right triangle.

Remember!

Holt Geometry


Given: Rectangle PQRS

Prove: The diagonals are .

Example 4: Writing a Coordinate Proof

The coordinates of P are (0, b),

the coordinates of Q are (a, b),

the coordinates of R are (a, 0),

and the coordinates of S are (0, 0).

Step 1 Assign coordinates to each vertex.

Step 2 Position the figure in the coordinate plane.

Holt Geometry


Given: Rectangle PQRS

Prove: The diagonals are .

Example 4 Continued

By the distance formula, PR = √ a2 + b2, and

QS = √a2 + b2 . Thus the diagonals are .

Step 3 Write a coordinate proof.

Holt Geometry



Use the information in Example 4 to write a coordinate proof showing that the area of ∆ADB is one half the area of ∆ABC.

The coordinates of A are (0, 2j),

the coordinates of B are (0, 0),

and the coordinates of C are (2n, 0).

Step 1 Assign coordinates to each vertex.

Step 2 Position the figure in the coordinate plane.

Holt Geometry


Step 3 Write a coordinate proof.


Holt Geometry


By the Midpoint Formula, the coordinates of

D = = (n, j). 0 + 2n, 2j + 0 2 2


Proof: ∆ABC is a right triangle with height 2j and base 2n.

The area of ∆ABC = bh

= (2n)(2j)

= 2nj square units

1212

Holt Geometry



The height of ∆ADB is j units, and the base is 2n units.

area of ∆ADB = bh

= (2n)(j)

= nj square units

Since nj = (2nj), the area of ∆ADB is one half the

area of ∆ABC.

12

1212

Holt Geometry


Lesson Quiz: Part I

Position each figure in the coordinate plane.

2. square with side lengths of 5a units

1. rectangle with a length of 6 units and a width of 3 units

Possible answers:

Holt Geometry


Lesson Quiz: Part II

3. Given: Rectangle ABCD with coordinates A(0, 0), B(0, 8), C(5, 8), and D(5, 0). E is mdpt. of BC, and F is mdpt. of AD.

Prove: EF = AB

By the Midpoint Formula, the coordinates of E are

, 8 .

and F are ,0 . Then EF = 8, and AB = 8.

Thus EF = AB.

52

52

Equations of Circles

Equation of a Circle

The center of a circle is given by (h, k)

The radius of a circle is given by r

The equation of a circle in standard form is (x – h)2 + (y – k)2 = r2

Finding the Equation of a Circle Circle A

The center is (16, 10)

The radius is 10

The equation is (x – 16)2 + (y – 10)2 = 100

Finding the Equation of a Circle

Circle B


The radius is 10

The equation is (x – 4)2 + (y – 20)2 = 100

Finding the Equation of a Circle Circle O


The radius is 12

The equation is x 2 + y 2 = 144

Graphing Circles

(x – 3)2 + (y – 2)2 = 9

Center (3, 2)

Radius of 3

Graphing Circles

(x + 4)2 + (y – 1)2 = 25

Center (-4, 1)

Radius of 5

Graphing Circles

(x – 5)2 + y2 = 36

Center (5, 0)

Radius of 6

Writing Equations of Circles

Write the standard equation of the circle:

Center (4, 7) Radius of 5

(x – 4)2 + (y – 7)2 = 25



Center (-3, 8) Radius of 6.2

(x + 3)2 + (y – 8)2 = 38.44



Center (2, -9) Radius of

(x – 2)2 + (y + 9)2 = 11

11



Center (0, 6) Radius of

x 2 + (y – 6)2 = 7

7



Center (-1.9, 8.7) Radius of 3

(x + 1.9)2 + (y – 8.7)2 = 9

The standard form of the equation of a circle with its center at the origin is

222 ryx

Notice that both the x and y terms are squared. Linear equations don’t have either the x or y terms squared. Parabolas have only the x term was squared (or only the y term, but NOT both).

r is the radius of the circle so if we take the square root of the right hand side, we'll know how big the radius is.

Let's look at the equation 922 yx

The center of the circle is at the origin and the radius is 3. Let's graph this circle.

This is r2 so r = 3

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8

Center at (0, 0)Count out 3 in all directions since that is the radius

If the center of the circle is NOT at the origin then the equation for the standard form of a circle looks like this:

222 rkyhx

The center of the circle is at (h, k).

1613 22 yx

The center of the circle is at (h, k) which is (3,1).

Find the center and radius and graph this circle.

The radius is 4

This is r2 so r = 4

2

-7

-6

-5

-4

-3

-2

-1

1

5

7

3

0

4

6

8

If you take the equation of a circle in standard form for example: 442 22 yx

You can find the center and radius easily. The center is at (-2, 4) and the radius is 2.

Remember center is at (h, k) with (x - h) and (y - k) since the x is plus something and not minus, (x + 2) can be written as (x - (-2))

This is r2 so r = 2

(x - (-2))

But what if it was not in standard form but multiplied out (FOILED)

416844 22 yyxxMoving everything to one side in descending order and combining like terms we'd have:

0168422 yxyx

0168422 yxyxIf we'd have started with it like this, we'd have to complete the square on both the x's and y's to get in standard form.

______16____8____4 22 yyxx

Group x terms and a place to complete the square

Group y terms and a place to complete the square

Move constant to the other side

4 416 16

442 22 yx

Write factored and wahlah! back in standard form.

Complete the square

Now let's work some examples:

Find an equation of the circle with center at (0, 0) and radius 7.

222 rkyhx

Let's sub in center and radius values in the standard form

0 0 7

4922 yx

Find an equation of the circle with center at (0, 0) that passes through the point (-1, -4).

222 ryx

222 41 r

The point (-1, -4) is on the circle so should work when we plug it in the equation:

17161

1722 yx

Since the center is at (0, 0) we'll have

Subbing this in for r2 we have:

Find an equation of the circle with center at (-2, 5) and radius 6

Subbing in the values in standard form we have:

222 rkyhx -2 5 6

3652 22 yx

Find an equation of the circle with center at (8, 2) and passes through the point (8, 0).

Subbing in the center values in standard form we have:

222 rkyhx 8 2

222 2088 r

Since it passes through the point (8, 0) we can plug this point in for x and y to find r2.

4

428 22 yx

Identify the center and radius and sketch the graph:

To get in standard form we don't want coefficients on the squared terms so let's divide everything by 9.

So the center is at (0, 0) and the radius is 8/3.

9

6422 yx

6499 22 yx9 9 9

Remember to square root this to get the radius.

2

-7

-6

-5

-4

-3

-2

-1

1

5

7

3

0

4

6

8

Identify the center and radius and sketch the graph:

Remember the center values end up being the opposite sign of what is with the x and y and the right hand side is the radius squared.

So the center is at (-4, 3) and the radius is 5.

2534 22 yx

2 -

7 -

6 -

5 -

4 -

3 -

2 -

1 1 5 7 3 0 4 6 8

034622 yxyx

We have to complete the square on both the x's and y's to get in standard form.

______3____4____6 22 yyxx

Group x terms and a place to complete the square

Group y terms and a place to complete the square

Move constant to the other side

9 94 4

1623 22 yx

Write factored for standard form.

Find the center and radius of the circle:

So the center is at (-3, 2) and the radius is 4.

Project of II-Prime

SY. 2011-2012

geometry 3rd-4th grading lessons

Education

corresponding segments

opposite angles arecongruent2

opposite sides arecongruent2

d ax cy

x dividessegment abso

ab cd c6 y

points b

y divides cy d