Worksheet 12 (3.1)

Chapter 3 Linear Equations and Inequalities in

Two Variables3.1 Rectangular Coordinate System

Summary 1:

Summary 2:

The Rectangular Coordinate System

A plane can be divided into four quadrants by two perpendicular number lines, called the horizontal and vertical axes. The horizontal axis is usually referred to as the x-axis and the vertical axis as the y-axis. The point of intersection of the two axes is called the origin. The quadrants are numbered counterclockwise I through IV beginning at the top right corner. Every point in the plane corresponds uniquely to an ordered pair of real numbers, called the coordinates of the point. Ordered pairs are listed in x, y order: (x,y) The first coordinate, x, called the abscissa, gives the location of the point, left or right, from the origin. The second coordinate, y, called the ordinate, gives the location of the point, up or down, from the origin. The ordered pair (-2,3) tells us that the point is located two units to the left of the origin and three units up from the origin. This point would be in Quadrant II. The coordinates of the origin would be the ordered pair (0,0). This system is called the rectangular coordinate system or the Cartesian coordinate system.

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Worksheet 12 (3.1)

Warm-up 1. a) Graph: x - 2y = 4

1. First calculate ordered pairs which are solutions for the equation:

Choose x

Find y Solutions for x - 2y = 4

2 2 - 2y = 4 y = -1

(2 , -1)

0 0 - 2y = 4 y =

(0 , )

4 ( ) - 2y = 4 y =

( , )

Note: Any values may be chosen for x, but choosing a "nice" value will make the computation easier.

2. Graphing the 3 ordered pairs obtained in the chart and drawing a line through the points will give the graph of the equation above.

y

x

Graphing an Equation To graph the equation y = x + 3, we would first find ordered pairs of numbers which are solutions for the equation. EX (2,5) is a solution for the equation because if we substitute 2 for x and 5 for y in the equation, we have a true statement: 5 = 2 + 3. There are an infinite number of points, represented by ordered pairs, that will satisfy the equation and lie on the graph of the line.

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The ordered pair (4,0) is called the x-intercept of the graph since this is the point where the graph crosses the x axis. (To easily find the x-intercept, let y = 0 and solve for x.)

The ordered pair ( , ) is called the y-intercept of the graph since this is the point where the graph crosses the axis. (To easily find the y-intercept, let x = and solve for .)

Worksheet 12 (3.1)Problem

1. Graph: 2x + y = 6 y

x

Summary 3:

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Symmetry The graphs of many equations exhibit the property of symmetry. Graphs may exhibit: 1. y-axis symmetry - Each half of the curve is a mirror image of the other half through the y-axis.

- If an ordered pair (x,y) is a solution for the equation, then the ordered pair (-x,y) will also be a solution. 2. x-axis symmetry - Each half of the curve is a mirror image of the

other half through the x-axis.- If an ordered pair (x,y) is a solution for

the equation, then the ordered pair (x,-y) will also be a solution. 3. origin symmetry - Each half of the curve is a mirror image of the other half through the origin.

- If an ordered pair (x,y) is a solution for the equation, then the ordered pair (-x,-y) will also be a solution.

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Warm-up 2. a) For the point (-5,2), determine the point that is symmetric with respect to the x-axis.

The point symmetric to (-5,2) with respect to the x-axis is (-5, ).

b) For the point (-5,2), determine the point that is symmetric with respect to the y-axis.

The point symmetric to (-5,2) with respect to the y-axis is ( , 2).

c) For the point (-5,2), determine the point that is symmetric with respect to the origin.

The point symmetric to (-5,2) with respect to the origin is ( , ).

Worksheet 12 (3.1)

Problems

2. For the point (7,-4), determine the point that is symmetric with respect to the

x-axis.

3. For the point (-3,-1), determine the point that is symmetric with respect to the

y-axis.

4. For the point (-6,2), determine the point that is symmetric with respect to the origin.

Summary 4:

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Warm-up 3. Determine the type of symmetry (y-axis, x-axis, or origin) exhibited by each of the graphs of the following equations:

a) y = 3x

Test for y-axis symmetry: replace x with -x. y = 3(-x) y = -3x (not an equivalent equation)

Test for x-axis symmetry: replace y with -y.

1. y-axis symmetry - The graph of an equation is symmetric with respect to the y-axis if replacing x with -x results in an equivalent equation.

2. x-axis symmetry - The graph of an equation is symmetric with respect to the x-axis if replacing y with -y results in an equivalent equation.

3. origin symmetry - The graph of an equation is symmetric with respect to the origin if replacing x with -x and y with -y results in an equivalent equation.

Tests for Symmetry

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-y = 3x (not an equivalent equation)

Test for origin symmetry: replace x with -x and y with -y.

-y = 3(-x) -y = -3x y = 3x (equivalent equation)

The graph is symmetric with respect to the .Worksheet 12 (3.1)

b) x = y2 - 4

Test for y-axis symmetry: replace with . -x = y2 - 4 (not an equivalent equation)

Test for x-axis symmetry: replace with . x = (-y)2 - 4 x = (equivalent equation)

The graph is symmetric with respect to the .

c) y = x4 - 16

Test for y-axis symmetry: replace with . y = (-x)4 - 16 y = (equivalent equation)

The line is symmetric with respect to the .

Problems - Determine the type of symmetry (y-axis, x-axis or origin) exhibited by each of the graphs of the following equations:

5. x = 9 - y2

6. xy = 4

7. y = x2 + 15

Summary 5:

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Worksheet 12 (3.1)

Warm-up 4. Graph:

a) y = x2 - 4

1. Determine the type of symmetry: y = x2 - 4 y = (-x)2 - 4 (replace x with -x) y = x2 - 4 (equivalent equation)

The equation has symmetry.

2. Find the y-intercept: y = (0)2 - 4 (let x = 0, solve for y) y =

The y-intercept is (0, ).

Find the x-intercepts: 0 = x2 - 4 (let y = 0, solve for x) 0 = (x - 2)(x + 2)

x = or x = The x-intercepts are ( , 0) and ( , 0).

3. The equation is already solved for y.

4. Set up a table of ordered pairs.

x y

2 0 x-

Graphing Suggestions 1. Determine the type of symmetry exhibited by the equation. 2. Find the x-intercept(s) and y-intercept(s), if they exist. 3. Solve the equation for y or x if it is not already in this form. 4. Set up a table of ordered pairs that satisfy the equation. 5. Plot the points from the table, connect with a smooth curve. If appropriate, reflect this part of the curve according to the symmetry shown in the equation.

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intercept-2 0 x-

intercept 0 -4 y-

intercept 1 -3-1

5. Plot the points, connect with a smooth curve.

y

x

Worksheet 12 (3.1)b)

1. Determine the type of symmetry: (replace x with -x and y with -y)

(equivalent equation)

The equation has symmetry.

2. In this equation x can not be 0 and y will never be 0, therefore there are no x-intercepts or y-intercepts.

3. The equation is already solved for y.

4. Set up a table of ordered pairs.

x y

1 ½ 2 ¼ ½ 1-1

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-2-½

5. Plot the points and connect with a smooth curve. The graph will have two identical halves, symmetric to the origin. The graph will be in Quadrant I and III. Since x and y can not

be zero, the graph will approach the axes, but never touch them.

y

x

Worksheet 12 (3.1)

Problems - Graph:

8. x = y3 + 1 y

x

9. x = 2y2 - 2 y

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x

Worksheet 13 (3.2)

3.2 Linear Equations in Two Variables

Summary 1:

Linear Equations in Two Variables

A linear equation in two variables is any equation of the form

Ax + By = C where A, B, C are constants and A and B not both = 0. The graph of this type of equation is a straight line.

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Warm-up 1. a) Graph: 2x + 4y = 8

x y

0 x-intercept 0 y-intercept 2 check

pointy

xb) Graph: y = 3x

x y

0 0 x and y intercepts

1 -1

check point

y

xThis equation exhibits origin symmetry.Note: Since the x and y intercepts on Warm-up 1b) were the same, we had to find another point in addition to the check point.

Worksheet 13 (3.2)

Problems - Graph:

1. 3x - 4y = 12 y

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x

2. y = -2x y

x

Summary 2:

Warm-up 2. a) Graph: y = 2 y

Horizontal and Vertical Lines

1. Horizontal lines have a standard form of 0x + y = c, or just y = c. For every x value, y will always equal the constant c.

Horizontal lines exhibit y-axis symmetry.

2. Vertical lines have a standard form of x + 0y = c, or just x = c.

For every y value, x will always equal the constant c.Vertical lines exhibit x-axis symmetry.

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x

This is an equation in standard form for a horizontal line. Draw a horizontal line through the y-intercept of ______. Note: This graph exhibits y-axis symmetry.

Worksheet 13 (3.2)

b) Graph: x = -4 y

x

This is an equation in standard form for a vertical line. Draw a vertical line through thex-intercept of . Note: This graph exhibits x-axis symmetry.

Problems - Graph:

3. x = 3 y

x

4. y = -1 y

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x

Worksheet 14 (3.3)

3.3 Linear Inequalities in Two Variables

Summary 1:

Summary 2:

Linear Inequalities

Linear inequalities in two variables are in the form:Ax + By > C or Ax + By < 0 (equality is not included)Ax + By C or Ax + By C (equality is included)

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Warm-up 1. a) Graph: 2x + y 6

1. Graph the x-intercept (3,0) and the y-intercept (0,6) of the equality. Draw a solid line through the two points.

( symbol includes equality).2. Test (0,0): 2(0) + 0 6

6 True or False?3. Shade the half-plane containing (0,0).

(statement was true)y

x

Worksheet 14 (3.3)

b) Graph: x > 2

1. Graph the corresponding equality, or boundary line. Use a solid line if equality is included in the original statement ( or ). Use a dashed line if equality is not included (< or >). 2. Choose a "test point" not on the line and check it in the inequality statement. (The point (0,0) is an easy test point if not on the line.) 3. If the test point satisfies the inequality, shade the half-plane containing the test point. If the test point does not satisfy the inequality, shade the half-plane that does not contain the test point.

Graphing Linear Inequalities

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1. Graph the equality x = 2 as a vertical line.

(< symbol does not have equality)

2. Test (0,0): 0 > 2 True or False?3. Shade the half-plane not containing (0,0). (statement was false)

y

x

Problems - Graph:

1. x - 2y > 4 y

x

2. y 4 y

x

Worksheet

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15 (3.4)

3.4 Distance and Slope

Summary 1:

Warm-up 1. a) Find the distance between the points P1 (-3,4) and P2 (5,1).

1. Label the points: (-3,4) = (x1, y1); (5,1) = (x2, y2)2. Write the distance formula:

3. Substitute in the values given.

Problem - Find the distance between the points:

1. P1 (4,-2) and P2 (0, -5)

The distance between 2 points P1 (x1, y1) and P2 (x2, y2) in the rectangular coordinate plane is found by the formula:

Distance Formula

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Worksheet 15 (3.4)

Summary 2:

Warm-up 2. Graph the line determined by the two points and find the slope of the line.

a) (-3,4) and (5,0)

1. Graph the two points and draw the line determined.

y

Slope

Slope is used to describe the "steepness" of a line. The slope of a line is the ratio of the vertical change compared to the horizontal change as we move from one point on a line to another.

The slope of the line through two points, P1(x1,y1) and P2(x2, y2) is found by the formula: , x2 x1

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x

2. Let (-3,4) = (x1, y1); (5,0) = (x2, y2)3. Write the slope formula:

4. Substitute in values:

Worksheet 15 (3.4)

b) (5, -2) and (3, -2)

1. Graph the two points and draw the line determined.

y

x

2. Let (5, -2) = (x1, y1) and (3, -2) = (x2, y2) 3. Substitute into slope formula:

Note: These 2 points are points on a line. (Flat line, slope = 0). When a line has a slope = 0, the line will be horizontal.

c) (3, -1) and (3, 4)

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1. Graph the two points and draw the line determined.

y

x

2. Let (3, -1) = (x1, y1) and (3, 4) = (x2, y2) 3. Substitute into slope formula:

= undefined

Note: The slope is undefined because division by zero is undefined. These 2 points are points on a line. (Line is straight up, slope cannot be determined.) When a line has a slope which is undefined, the line will be vertical. Worksheet 15 (3.4)

Problems - Find the slope of the line containing the two points and graph the line determined by the points.

2. (2, -1) and (-4, 3) y

x

3. (2, -5) and (2, 3) y

x

4. (-1, 4) and (2, 4) y

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xSummary 3:

Worksheet 15 (3.4)

Warm-up 3. a) A staircase is to have a ratio of rise to run of ½ (slope = ½). If the tread of the steps (run) is to be 18 inches, find the rise.

Set up a proportion

2x = x =

1. Slope can be positive. The line will rise as you move from left to right. (Line tilts to the right.) 2. Slope can be negative. The line will fall as you move from left to right. (Line tilts to the left.) 3. Slope can equal 0. The line will be horizontal and does not rise or fall as you move from left to right. 4. Slope can be undefined. The line will be vertical, and does not move from left to right.

Possibilities for Slope

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The rise of the stairs is when the run is 18 inches.

Problem

5. The pitch or slope of the roof of an A-frame house is 2 (write as 2/1). If the rise is 24 inches find the run.

Worksheet 16 (3.5)

3.5 Determining the Equation of a Line

Summary 1:

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Warm-up 1. a) Find the equation of the line that has a slope of -¾ and contains the point (-2, 4).

1. Identify the slope: m = -¾

2. Identify a point: (x1, y1) = (-2, 4)

3. Begin with point-slope form:

y - y1 = m(x - x1) y - = (x - ( )) substitute in m, x1,

y1 y - = (x + )

(y - 4) = (-¾)(x + 2) multiply by LCD 4y - 16 = -3(x + 2) 4y - 16 = - -

4. Rewrite the equation in standard form:

+ =

b) Find the equation of the line containing the points (-2, 4) and (1, -3).

1. Identify the slope:

Point-Slope Form Point-slope form:

y - y1 = m(x - x1) where m = slope and (x1, y1) = any point on the line.

Given certain facts about a line we can determine its algebraic equation. When a point is given and the slope is either given or enough information is given to find the slope, we can use the point-slope form to write the equation of the line.

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Worksheet

16 (3.5)

2. Identify a point: (x1, y1) = (-2, 4)

3. Begin with point-slope form:

y - = (x - ) 3(y - ) = 3( )(x + ) 3y - = -7(x + ) 3y - = -7x -

4. Rewrite in standard form:

+ =

Problems

1. Find the equation of the line that has a slope of 4 and contains the point (3, -5). Write your final answer in standard form.

2. Find the equation of the line that contains the points (-3, 3) and (1, -5). Write your final answer in standard form.

Summary 2:

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Worksheet 16 (3.5)

Warm-up 2. a) Find the equation of the line that has a slope = -½ and y-intercept of 3. Write the final equation in slope-intercept form.

1. Identify the slope: m =

2. Identify the y-intercept: b =

3. Use slope-intercept form:

y = mx + by = x + Substitute in m and b

Problem

3. Find the equation of the line that has a slope = 2/5 and a y-intercept of -5.

Warm-up 3. a) Find the slope of the line whose equation is 2x - y = 5.

Note: If the equation is rewritten in slope-intercept form we can identify the slope and y-intercept.

When the slope and the y-intercept of a line are given we use the slope- intercept form to find the equation of the line:

y = mx + b where m = slope and b = y-intercept

Slope-Intercept Form

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2x - y = 5 - y = + 5 y = - 5

Slope =

Problem

4. Find the slope of the line whose equation is x + 3y = 6.

Worksheet 16 (3.5)

Summary 3:

Parallel and Perpendicular Lines

1. Two lines are parallel if and only if their slopes are equal. (m1 = m2)

2. Two lines are perpendicular if and only if their slopes are negative reciprocals. (If m1 = 2/3 then m2 = -3/2)

Note: This can also be stated, the product of the 2 slopes equals -1. (m1)(m2) = -1

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Warm-up 4. a) Write the equation of a line that contains the point (-3, 5) and is perpendicular to the line 2x - 3y = 9.

1. Identify the slope: Since the line is perpendicular to

2x - 3y = 9, the slope of the line will be the negative reciprocal of the slope of 2x - 3y = 9.

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

Slope of this line is , slope of the line perpendicular to it would be .

2. Identify a point: (x1, y1) = (-3, 5)

3. Use point-slope form:

y - 5 = (x - ( )) 2(y - 5) = 2( )(x + ) 2y - = -3(x + ) 2y - = -3x -

4. Rewrite in standard form: + =

Problem

5. Write the equation of the line that contains the point (2, -6) and is parallel to the line 4x - 2y = 6.

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