Coordinate Graphs of Quadrilaterals

The illustrated problems given below are graphs of the following quadrilaterals:

• rectangles
• squares
• rhombi
• trapezoids

A quadrilateral is defined as a polygon with four sides.

A parallelogram is a quadrilateral that has two pairs of parallel opposite sides.

Description of Quadrilaterals

Rectangles and rhombi are both parallelograms. However, rhombi have no right interior angles while rectangles have right interior angles.

Generally, a rectangle has one of its two consecutive sides longer than the other one. But a square, which is also a rectangle, has two equal consecutive sides.

Like the square, a rhombus has two equal consecutive sides. But it has no right interior angles.

Lastly, a trapezoid is a quadrilateral with one pair of parallel opposite sides. Its non-parallel sides can be equal or unequal. If a trapezoid has equal non-parallel sides, then it's called an isosceles trapezoid.

Concepts Illustrated By the Problems

Each problem below illustrates the following concepts:

• distance between 2 points
• parallel and perpendicular lines
• point of intersection of two straight lines
• equation of a straight line

The equation of a straight line can be determined if any of the following are given:

• 2 points on the line
• slope and a point of the line
• slope and the x-intercept ( or y-intercept ) of the line
• x-intercept and y-intercept of the line

Description of the Problems

For each problem (or graph), the following information are given:

• equations of the line segments
• slopes of the line segments
• x-intercepts and y-intercepts of the lines
• distances between the terminal points of the line segments
• points on the extended line segments

Using one or a combination of two or more of the above information, it is possible for a problem to be presented in several ways.

Line segments and their terminal points referred to in the problems are the sides and vertices, respectively, of the quadrilaterals.

Slope of a line segment with terminal points A and B is abbreviated as m_AB.

Distance between points A and B is abbreviated as distance_AB

Points on a line refer to the points of the extended line of that line segment.

Graphs of Rectangles

Show that:

• points A(-2, 4), B(-4, 0), C(0, 3), and D(-2, -1) are vertices of a rectangle;
• points A(-2, 4), B(-4, 0), E(4, 1), and F(2, -3) are vertices of a rectangle;
• points C(0, 3), D(-2, -1), E(4, 1), and F(2, -3) are vertices of a square.

Equations of the Line Segments

• line segment AB (2x - y + 8 = 0): x-intercept = -4; y-intercept = 8
• line segment CD (2x - y + 3 = 0): x-intercept = -3/2; y-intercept = 3
• line segment EF (2x - y - 7 = 0): x-intercept = 7/2; y-intercept = -7
• line segment AC or CE or AE (x + 2y - 6 = 0): x-intercept = 6; y-intercept = 3
• line segment BD or DF or BF (x + 2y + 4 = 0): x-intercept = -4; y-intercept = -2

Slopes of the Line Segments

• m_AB = m_CD = m_EF = 2
• m_AC = m_CE = m_AE = - 1/2
• m_BD = m_DF = m_BF = - 1/2

Distances Between Points

• distance_AB = distance_CD = distance_EF = √ 20
• distance_AC = distance_BD = √ 5
• distance_CE = distance_DF = √ 20
• distance_AE = distance_BF = √ 45

Points On The Lines

• line 2x - y + 8 = 0 (line segment AB): (-5, -2), (-4, 0), (-3, 2), (-2, 4), (-1, 6), (1, 10),(2, 12), (3, 14), (4, 16), (5, 18)
• line 2x - y + 3 = 0 (line segment CD): (-5, -7), (-4, -5), (-3, -3), (-2, -1), (-1, 1), (1, 5), (2, 7), (3, 9), (4, 11), (5, 13)
• line 2x - y - 7 = 0 (line segment EF): (-5, -17), (-4, -15), (-3, -13), (-2, -11), (-1, -9), (1, -5), (2, -3), (3, -1), (4, 1), (5, 3)
• line x + 2y - 6 = 0 (line segment AC or CE or AE): (-5, 11/2), (-4, 5), (-3, 9/2), (-2, 4), (-1, 7/2), (1, 5/2), (2, 2), (3, 3/2), (4, 1), (5, 1/2)
• line x + 2y + 4 = 0 (line segment BD or DF or BF): (-5, 1/2), (-4, 0), (-3, -1/2), (-2, -1), (-1, -3/2), (1, -5/2), (2, -3), (3, -7/2), (4, -4), (5, -9/2)

Show that the quadrilateral bounded by the lines 3x - 2y + 8 = 0 and 3x - 2y - 5 = 0 and line segments AB and CD is a rectangle.



Equations of the Line Segments

• line 3x - 2y + 8 = 0: x-intercept = - 8/3; y-intercept = 4
• line 3x - 2y - 5 = 0: x-intercept = 5/3; y-intercept = - 5/2
• line segment AB (2x + 3y - 12 = 0): x-intercept = 6; y-intercept = 4
• line segment CD (2x + 3y + 14 = 0): x-intercept = -7; y-intercept = -14/3

Slopes of the Line Segments

• m_AB = m_CD = - 2/3
• m_AD = m_BC = 3/2

Distances Between Points

• distance_AD = distance_BC = √ 52
• distance_AB = distance_CD = √ 13

Points On The Lines

• line 3x - 2y + 8 = 0 (line segment AD): (-5, -7/2), (-4, -2), (-3, -1/2), (-2, 1), (-1, 5/2), (1, 11/2), (2, 7), (3, 17/2), (4, 10), (5, 23/2)
• line 3x - 2y - 5 = 0 (line segment BC): (-5, -10), (-4, -17/2), (-3, -7), (-2, -11/2), (-1, -4), (1, -1), (2, 1/2), (3, 2), (4, 7/2), (5, 5)
• line 2x + 3y - 12 = 0 (line segment AB): (-5, 22/3), (-4, 20/3), (-3, 6), (-2, 16/3), (-1, 14/3), (1, 10/3), (2, 8/3), (3, 2), (4, 4/3), (5, 2/3)
• line 2x + 3y + 14 = 0 (line segment CD): (-5, -4/3), (-4, -2), (-3, -8/3), (-2, -10/3), (-1, -4), (1, -16/3), (2, -6), (3, -20/3), (4, -22/3), (5, -8)

Prove analytically that points A(-5, 2), B(3, 4), C(4, 0) and D(-4, -2) are vertices of a rectangle.



Equations of the Line Segments

• line segment AB : x - 4y + 13 = 0; x-intercept = -13; y-intercept = 13/4
• line segment BC : 4x + y - 16 = 0; x-intercept = 4; y-intercept = 16
• line segment CD : x - 4y - 4 = 0; x-intercept = 4; y-intercept = - 1
• line segment AD : 4x + y + 18 = 0; x-intercept = - 9/2; y-intercept = -18

Slopes of the Line Segments

• m_AB = m_CD = 1/4
• m_AD = m_BC = - 4

Distances Between Points

• distance_AB = distance_CD = √ 68
• distance_BC = distance_AD = √ 17

Points On The Lines

• line x - 4y + 13 = 0 (line segment AB): (-5, 2), (-4, 9/4), (-3, 5/2), (-2, 11/4), (-1, 3), (1, 7/2), (2,15/4 ), (3, 4), (4, 17/4), (5, 9/2)
• line 4x + y - 16 = 0 (line segment BC): (-5, 36), (-4, 32), (-3, 28), (-2, 24), (-1, 20), (1, 12), (2, 8), (3, 4), (4, 0), (5, -4)
• line x - 4y - 4 = 0 (line segment CD): (-5, -9/4), (-4, -2), (-3, -7/4), (-2, -3/2), (-1, -5/4), (1, -3/4), (2, -1/2), (3, -1/4), (4, 0), (5, 1/4)
• line 4x + y + 18 = 0 (line segment AD): (-5, 2), (-4, -2), (-3, -6), (-2, -10), (-1, -14), (1, -22), (2, -26), (3, -30), (4, -34), (5, -38)

Graphs of Squares

Show that points A(-1, 3), B(3, 2), C(2, -2) and D(-2, -1) are vertices of a square.



Equations of the Line Segments

• line segment AB : x + 4y - 11 = 0; x-intercept = 11; y-intercept = 11/4
• line segment BC : 4x - y - 10 = 0; x-intercept = 5/2; y-intercept = -10
• line segment CD : x + 4y + 6 = 0; x-intercept = -6; y-intercept = -3/2
• line segment AD : 4x - y + 7 = 0; x-intercept = -7/4; y-intercept = 7

Slopes of the Line Segments

• m_AB = m_CD = - 1/4
• m_BC = m_AD = 4

Distances Between Points

• distance_AB = distance_BC = distance_CD = distance_AD = √17

Points On The Lines

• line x + 4y - 11 = 0 (line segment AB): (-5, 4), (-4, 15/4), (-3, 7/2), (-2, 13/4), (-1, 3), (1, 5/2), (2, 9/4), (3, 2), (4, 7/4), (5, 3/2)
• line 4x - y - 10 = 0 (line segment BC): (-5, -30), (-4, -26), (-3, -22), (-2, -18), (-1, -14), (1, -6), (2, -2), (3, 2), (4, 6), (5, 10)
• line x + 4y + 6 = 0 (line segment CD): (-5, -1/4), (-4, -1/2), (-3, -3/4), (-2,-1 ), (-1, -5/4), (1, -7/4), (2, -2), (3, -9/4), (4, -5/2), (5, -11/4)
• line 4x - y + 7 = 0 (line segment AD): (-5, -13), (-4, -9), (-3, -5), (-2,-1 ), (-1, 3), (1, 11), (2, 15), (3, 19), (4, 23), (5, 27)

Show that the points of intersection of the following lines are vertices of a square:

• x + y - 4 = 0
• x - y - 4 = 0
• x + y + 4 = 0
• x - y + 4 = 0

Equations of the Line Segments

• line segment AB : x + y - 4 = 0; x-intercept = 4; y-intercept = 4
• line segment BC : x - y - 4 = 0; x-intercept = 4; y-intercept = -4
• line segment CD : x + y + 4 = 0; x-intercept = -4; y-intercept = -4
• line segment AD : x - y + 4 = 0; x-intercept = -4; y-intercept = 4

Slopes of the Line Segments

• m_AB = m_CD = - 1
• m_BC = m_AD = 1

Distances Between Points

• distance_AB = distance_BC = distance_CD = distance_AD = √ 32

Points On The Lines

• line x + y - 4 = 0 (line segment AB): (-5, 9), (-4, 8), (-3, 7), (-2, 6), (-1, 5), (1, 3), (2, 2), (3, 1), (4, 0), (5, -1)
• line x - y - 4 = 0 (line segment BC): (-5, -9), (-4, -8), (-3, -7), (-2, -6), (-1, -5), (1, -3), (2, -2), (3, -1), (4, 0), (5, 1)
• line x + y + 4 = 0 (line segment CD): (-5, 1), (-4, 0), (-3, -1), (-2, -2), (-1, -3), (1, -5), (2, -6), (3, -7), (4, -8), (5, -9)
• line x - y + 4 = 0 (line segment AD): (-5, -1), (-4, 0), (-3, 1), (-2, 2), (-1, 3), (1, 5), (2, 6), (3, 7), (4, 8), (5, 9)

Show that the points A(1, 5), B(3, 2), C(0, 0), and D(-2, 3) are vertices of a square.



Equations of the Line Segments

• line segment AB : 3x + 2y - 13 = 0; x-intercept = 13/3; y-intercept = 13/2
• line segment BC : 2x - 3y = 0; x-intercept = 0; y-intercept = 0
• line segment CD : 3x + 2y = 0; x-intercept = 0; y-intercept = 0
• line segment AD : 2x - 3y + 13 = 0; x-intercept = -13/2; y-intercept = 13/3

Slopes of the Line Segments

• m_AB = m_CD = - 3/2
• m_BC = m_AD = 2/3

Distances Between Points

• distance_AB = distance_BC = distance_CD = distance_AD = √ 13

Points On The Lines

• line 3x + 2y - 13 = 0 (line segment AB): (-5, 14), (-4, 25/2), (-3, 11), (-2, 19/2), (-1, 8), (1, 5), (2, 7/2), (3, 2), (4, 1/2), (5, -1)
• line 2x - 3y = 0 (line segment BC): (-5, -10/3), (-4, -8/3), (-3, -2), (-2, -4/3), (-1, -2/3), (1, 2/3), (2, 4/3), (3, 2), (4, 8/3), (5, 10/3)
• line 3x + 2y = 0 (line segment CD): (-5, 15/2), (-4, 6), (-3, 9/2), (-2, 3), (-1, 3/2), (1, -3/2), (2, -3), (3, -9/2), (4, -6), (5, -15/2)
• line 2x - 3y + 13 = 0 (line segment AD): (-5, 1), (-4, -5/3), (-3, 7/3), (-2, 3), (-1, 11/3), (1, 5), (2, 17/3), (3, 19/3), (4, 7), (5, 23/3)

Graphs of Rhombi

Prove that points A(1, 5), B(4, 1), C(1, -3) and D(-2, 1) are vertices of a rhombus.



Equations of the Line Segments

• line segment AB : 4x + 3y - 19 = 0; x-intercept = 19/4; y-intercept = 19/3
• line segment BC : 4x - 3y - 13 = 0; x-intercept = 13/4; y-intercept = - 13/3
• line segment CD : 4x + 3y + 5 = 0; x-intercept = - 5/4; y-intercept = - 5/3
• line segment AD : 4x - 3y + 11 = 0; x-intercept = - 11/4; y-intercept = 11/3

Slopes of the Line Segments

• m_AB = m_CD = - 4/3
• m_BC = m_AD = 4/3

Distances Between Points

• distance_AB = distance_BC = distance_CD = distance_AD = √ 25

Points On The Lines

• line 4x + 3y - 19 = 0 (line segment AB): (-5, 13), (-4, 35/3), (-3, 31/3), (-2, 9), (-1, 23/3), (1, 5), (2, 11/3), (3, 7/3), (4, 1), (5, -1/3)
• line 4x - 3y - 13 = 0 (line segment BC): (-5, -11), (-4, -29/3), (-3, -25/3), (-2, -7), (-1, -17/3), (1, -3), (2, -5/3), (3, -1/3), (4, 1), (5, 7/3)
• line 4x + 3y + 5 = 0 (line segment CD): (-5, 5), (-4, 11/3), (-3, 7/3), (-2, 1), (-1, -1/3), (1, -3), (2, -13/3), (3, -17/3), (4, -7), (5, -25/3)
• line 4x - 3y + 11 = 0 (line segment AD): (-5, -3), (-4, -5/3), (-3, -1/3), (-2, 1), (-1, 7/3), (1, 5), (2, 19/3), (3, 23/3), (4, 9), (5, 31/3)

Show that points A(-2, 3), B(3, 4), C(2, -1), and D(-3, -2) are vertices of a rhombus.



Equations of the Line Segments

• line segment AB : x - 5y + 17 = 0; x-intercept = -17; y-intercept = 17/5
• line segment BC : 5x - y - 11 = 0; x-intercept = 11/5; y-intercept = -11
• line segment CD : x - 5y - 7 = 0; x-intercept = 7; y-intercept = - 7/5
• line segment AD : 5x - y + 13 = 0; x-intercept = - 13/5; y-intercept = 13

Slopes of the Line Segments

• m_AB = m_CD = 1/5
• m_BC = m_AD = 5

Distances Between Points

• distance_AB = distance_BC = distance_CD = distance_AD = √ 26

Points On The Lines

• line x - 5y + 17 = 0 (line segment AB): (-5, 12/5), (-4, 13/5), (-3, 14/5), (-2, 3), (-1, 16/5), (1, 18/5), (2, 19/5), (3, 4), (4, 21/5), (5, 22/5)
• line 5x - y - 11 = 0 (line segment BC): (-5, -36), (-4, -31), (-3, -26), (-2, -21), (-1, -16), (1, -6), (2, -1), (3, 4), (4, 9), (5, 14)
• line x - 5y - 7 = 0 (line segment CD): (-5, -12/5), (-4, -11/5), (-3, -2), (-2, -9/5), (-1, 8/5), (1, -6/5), (2, -1), (3, -4/5), (4, -3/5), (5, -2/5)
• line 5x - y + 13 = 0 (line segment AD): (-5, -12), (-4, -7), (-3, -2), (-2, 3), (-1, 8), (1, 18), (2, 23), (3, 28), (4, 33), (5, 38)

Prove that points A(-5, 2), B(3/5, 4), C(5, 0), and D(-3/5, -2) are vertices of a rhombus.



Equations of the Line Segments

• line segment AB : 5x - 14y + 53 = 0; x-intercept = - 53/5; y-intercept = 53/14
• line segment BC : 10x + 11y - 50 = 0; x-intercept = 5; y-intercept = 50/11
• line segment CD : 5x - 14y - 25 = 0; x-intercept = 5; y-intercept = - 25/14
• line segment AD : 10x + 11y + 28 = 0; x-intercept = - 14/5; y-intercept = - 28/11

Slopes of the Line Segments

• m_AB = m_CD = 5/14
• m_BC = m_AD = - 10/11

Distances Between Points

• distance_AB = distance_BC = distance_CD = distance_AD = √ 884/25

Points On The Lines

• line 5x - 14y + 53 = 0 (line segment AB): (-5, 2), (-4, 33/14), (-3, 19/7), (-2, 43/14), (-1, 24/7), (1, 29/7), (2, 9/2), (3, 34/7), (4, 73/14), (5, 39/7)
• line 10x + 11y - 50 = 0 (line segment BC): (-5, 100/11), (-4, 90/11), (-3, 80/11), (-2, 70/11), (-1, 60/11), (0, 50/11), (1, 40/11), (2, 30/11), (3, 20/11), (4, 10/11), (5, 0)
• line 5x - 14y - 25 = 0 (line segment CD): (-5, 25/7), (-4, -45/14), (-3, -20/7), (-2, -5/2), (-1, -15/7), (1, -10/7), (2, -15/14), (3, -5/7), (4, -5/14), (5, 0)
• line 10x + 11y + 28 = 0 (line segment AD): (-5, 2), (-4, 12/11), (-3, 2/11), (-2, -8/11), (-1, -18/11), (1, -38/11), (2, -48/11), (3, -58/11), (4, -68/11), (5, -78/11)

Graphs of Trapezoids

Prove that the points A(-1, 4), B(4, 5), C(-1, -5) and D(-4, -2) are vertices of a trapezoid.



Equations of the Line Segments

• line segment AB : x - 5y + 21 = 0; x-intercept = - 21; y-intercept = 21/5
• line segment BC : 2x - y - 3 = 0; x-intercept = 3/2; y-intercept = - 3
• line segment CD : x + y + 6 = 0; x-intercept = - 6; y-intercept = - 6
• line segment AD : 2x - y + 6 = 0; x-intercept = -3; y-intercept = 6

Slopes of the Line Segments

• m_AB = 1/5
• m_CD = -1
• m_BC = m_AD = 2

Distances Between Points

• distance_AB: √ 26
• distance_BC: √ 125
• distance_CD: √ 18
• distance_AD: √ 45

Points On The Lines

• line x - 5y + 21 = 0 (line segment AB): (-5, 16/5), (-4, 17/5), (-3, 18/5), (-2, 19/5), (-1, 4), (1, 22/5), (2, 23/5), (3, 24/5), (4, 5), (5, 26/5)
• line 2x - y - 3 = 0 (line segment BC): (-5, -13), (-4, -11), (-3, -9), (-2, -7), (-1, -5), (1, -1), (2, 1), (3, 3), (4, 5), (5, 7)
• line x + y + 6 = 0 (line segment CD): (-5, -1), (-4, -2), (-3, -3), (-2, -4), (-1, -5), (1, -7), (2, -8), (3, -9), (4, -10), (5, -11)
• line 2x - y + 6 = 0 (line segment AD): (-5, -4), (-4, -2), (-3, 0), (-2, 2), (-1, 4), (1, 8), (2, 10), (3, 12), (4, 14), (5, 16)

Show that points A(- 80/17, 71/17), B(-1, 5), C(2, 0), and D(- 8/17, - 49/17) are vertices of an isosceles trapezoid.



Equations of the Line Segments

• line segment AB : 2x - 9y + 47 = 0; x-intercept = - 47/2; y-intercept = 47/9
• line segment BC : 5x + 3y - 10 = 0; x-intercept = 2; y-intercept = 10/3
• line segment CD : 7x - 6y - 14 = 0; x-intercept = 2; y-intercept = - 7/3
• line segment AD : 5x + 3y + 11 = 0; x-intercept = - 11/5; y-intercept = - 11/3

Slopes of the Line Segments

• m_AB = 2/9
• m_CD = 7/6
• m_BC = m_AD = - 5/3

Distances Between Points

• distance_AB = distance_CD = √ 245/17
• distance_BC = √ 34
• distance_AD = √ 1152/17

Points On The Lines

• line 2x - 9y + 47 = 0 (line segment AB): (-5, 37/9), (-4, 13/3), (-3, 41/9), (-2, 43/9), (-1, 5), (1, 49/9), (2, 17/3), (3, 53/9), (4, 55/9), (5, 19/3)
• line 5x + 3y - 10 = 0 (line segment BC): (-5, 35/3), (-4, 10), (-3, 25/3), (-2, 20/3), (-1, 5), (1, 5/3), (2, 0), (3, -5/3), (4, -10/3), (5, -5)
• line 7x - 6y - 14 = 0 (line segment CD): (-5, - 49/6), (-4, -7), (-3, - 35/6), (-2, - 14/3), (-1, - 7/2), (1, - 7/6), (2, 0), (3, 7/6), (4, 7/3), (5, 7/2)
• line 5x + 3y + 11 = 0 (line segment AD): (-5, 14/3), (-4, 3), (-3, 4/3), (-2, -1/3), (-1, -2), (1, -16/3), (2, -7), (3, -26/3), (4, -31/3), (5, -12)

Prove that points A(-4, 1), B(-1, 4), C(5, 1) and D(-1, -5) are vertices of an isosceles trapezoid.



Equations of the Line Segments

• line segment AB : x - y + 5 = 0; x-intercept = - 5; y-intercept = 5
• line segment BC : x + 2y - 7 = 0; x-intercept = 7; y-intercept = 7/2
• line segment CD : x - y - 4 = 0; x-intercept = 4; y-intercept = - 4
• line segment AD : 2x + y + 7 = 0; x-intercept = - 7/2; y-intercept = - 7

Slopes of the Line Segments

• m_AB = m_CD = 1
• m_BC = - 1/2
• m_AD = -2

Distances Between Points

• distance_AB = √ 18
• distance_CD = √ 72
• distance_BC = distance_AD = √ 45

Points On The Lines

• line x - y + 5 = 0 (line segment AB): (-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)
• line x + 2y - 7 = 0 (line segment BC): (-5, 6), (-4, 11/2), (-3, 5), (-2, 9/2), (-1, 4), (1, 3), (2, 5/2), (3, 2), (4, 3/2), (5, 1)
• line x - y - 4 = 0 (line segment CD): (-5, -9), (-4, -8), (-3, -7), (-2, -6), (-1, -5), (1, -3), (2, -2), (3, -1), (4, 0), (5, 1)
• line 2x + y + 7 = 0 (line segment AD): (-5, 3), (-4, 1), (-3, -1), (-2, -3), (-1, -5), (1, -9), (2, -11), (3, -13), (4, -15), (5, -17)

Algebra: Root of a Linear Equation In One Unknown (Fractional)

Analytic Geometry: Graphs of Intersecting And Parallel Lines

The following illustrated problems are about:

• intersecting lines (perpendicular and non-perpendicular)
• parallel lines

All the calculated values are given so as to allow flexibility in presenting the problems in different ways.

1-1 Graphs of Perpendicular Bisector of a Line Segment

Find the point at which the line x - 2y + 3 = 0 bisects perpendicularly the line segment formed by the x-intercept and y-intercept of the line 2x + y - 4 = 0.

• point of intersection (midpoint of line segment AB): (1,2)
• slope of the line x - 2y + 3 = 0: m = 1/2
• angle of inclination of line x - 2y + 3 = 0: 26.6°
• points on the line x - 2y + 3 = 0: {(-5, -1), (-4, -1/2), (-2, 1/2), (-1, 1), (0, 3/2), (-3, 0), (1, 2), (2, 5/2), (3, 3), (4, 7/2), (5, 4)}
• slope of the line 2x + y - 4 = 0: m = -2
• angle of inclination of line 2x + y - 4 = 0: 116.6°
• points on the line 2x + y - 4 = 0: {(-5, 14), (-4, 12), (-3, 10), (-2, 8), (-1, 6), (0, 4), (2, 0), (3, -2), (4, -4), (5, -6)}

Find the point at which the line x + 5y + 12 = 0 bisects perpendicularly the line segment formed by the x-intercept and y-intercept of the line 5x - y - 5 = 0.

• point of intersection (midpoint of line segment AB): (1/2, -5/2)
• slope of the line x + 5y + 12 = 0: m = -1/5
• angle of inclination of line x + 5y + 12 = 0: 168.7°
• points on the line x + 5y + 12 = 0: {(-5, -7/5), (-4, -8/5), (-3, -9/5), (-2, -2), (-1, -11/5), (0, -12/5), (-12, 0), (1, -13/5), (2, -14/5), (3, -3), (4, -16/5), (5, -17/5)}
• slope of the line 5x - y - 5 = 0: m = 5
• angle of inclination of line 5x - y - 5 = 0: 78.7°
• points on the line 5x - y - 5 = 0: {(-5, -30), (-4, -25), (-3, -20), (-2, -15), (-1, -10), (0, -5), (1, 0), (2, 5), (3, 10), (4, 15), (5, 20)}

Find the point at which the line 5x + 3y = 0 bisects perpendicularly the line segment AB.

• point of intersection (midpoint of line segment AB): (-3/2, 5/2)
• slope of the line 5x + 3y = 0: m = -5/3
• angle of inclination of line 5x + 3y = 0: 120.96°
• points on the line 5x + 3y = 0: {(-5, 25/3), (-4, 20/3), (-3, 5), (-2, 10/3), (-1, 5/3), (0, 0), (1, -5/3), (2, -10/3), (3, -5), (4, -20/3), (5, -25/3)}
• slope of the line segment AB: m = 3/5
• equation of line segment AB: 3x - 5y + 17 = 0

1-2 Graphs of Perpendicular Lines

Find the point of intersection of lines 8x + 5y + 13 = 0 and 5x - 8y - 3 = 0.

• point of intersection: (-1, -1)
• slope of the line 8x + 5y + 13 = 0: m = -8/5
• angle of inclination of line 8x + 5y + 13 = 0: 122.0°
• points on the line 8x + 5y + 13 = 0: {(-5, 27/5), (-4, 19/5), (-3, 11/5), (-2, 3/5), (0, -13/5), (-13/8, 0), (1, -21/5) (2, -29/5), (3, -37/5), (4, -9), (5, -53/5)}
• slope of the line 5x - 8y - 3 = 0: m = 5/8
• angle of inclination of line 5x - 8y - 3 = 0: 32.0°
• points on the line 5x - 8y - 3 = 0: {(-5, -7/2), (-4, -23/8), (-3, -9/4), (-2, -13/8), (0, -3/8), (3/5, 0), (1, 1/4), (2, 7/8), (3, 3/2), (4, 17/8), (5, 11/4))}

Find the point of intersection of lines x + 5y - 5 = 0 and 5x - y - 2 = 0.

• point of intersection: (15/26, 23/26)
• slope of the line x + 5y - 5 = 0: m = -1/5
• angle of inclination of line x + 5y - 5 = 0: 168.7°
• points on the line x + 5y - 5 = 0: {(-5, 2), (-4, 9/5), (-3, 8/5), (-2, 7/5), (-1, 6/5), (0, 1), (5, 0), (1, 4/5), (2, 3/5), (3, 2/5), (4, 1/5)}
• slope of the line 5x - y - 2 = 0: m = 5
• angle of inclination of line 5x - y - 2 = 0: 78.7°
• points on the line 5x - y - 2 = 0: {(-5, -27), (-4, -22), (-3, -17), (-2, -12), (-1, -7), (0, -2), (2/5, 0), (1, 3), (2, 8), (3, 13), (4, 18), (5, 23)}

1-3 Graphs of Intersecting Non-Perpendicular Lines

Find the point of intersection of lines 5x + 6y - 8 = 0 and 2x + y + 1 = 0.

• point of intersection: (-2, 3)
• slope of the line 5x + 6y - 8 = 0: m = -5/6
• angle of inclination of line 5x + 6y - 8 = 0: 140.2°
• points on the line 5x + 6y - 8 = 0: {(-5, 11/2), (-4, 14/3), (-3, 23/6), (-1, 13/6), (0, 4/3), (8/5, 0), (1, 1/2), (2, -1/3), (3, -7/6), (4, -2), (5, -17/6)}
• slope of the line 2x + y + 1 = 0: m = -2
• angle of inclination of line 2x + y + 1 = 0: 116.6°
• points on the line 2x + y + 1 = 0: {(-5, 9), (-4, 7), (-3, 5), (-1, 1), (0, -1), (-1/2, 0), (1, -3), (2, -5), (3, -7), (4, -9), (5, -11)}

Find the point of intersection of lines 5x + y + 16 = 0 and 3x - y + 8 = 0.

• point of intersection: (-3, -1)
• slope of the line 5x + y + 16 = 0: m = -5
• angle of inclination of line 5x + y + 16 = 0: 101.3°
• points on the line 5x + y + 16 = 0: {(-5, 9), (-4, 4), (-2, -6), (-1, -11), (0, -16), (-16/5, 0), (1, -21), (2, -26), (3, -31), (4, -36), (5, -41)}
• slope of the line 3x - y + 8 = 0: m = 3
• angle of inclination of line 3x - y + 8 = 0: 71.6°
• points on the line 3x - y + 8 = 0: {(-5, -7), (-4, -4), (-2, 2), (-1, 5), (0, 8), (-8/3, 0), (1, 11), (2, 14), (3, 17), (4, 20), (5, 23)}

2 Graphs of Parallel Lines

Show that lines 3x + 5y - 4 = 0 and 3x + 5y + 15 = 0 are parallel.

• points on the line 3x + 5y - 4 = 0: {(-5, 19/5), (-4, 16/5), (-3, 13/5), (-2, 2), (-1, 7/5), (0, 4/5), (4/3, 0), (1, 1/5), (2, -2/5), (3, -1), (4, -8/5), (5, -11/5)}
• slope of the lines: m = -3/5
• angle of inclination of the lines: 149.0°
• points on the line 3x + 5y + 15 = 0: {(-5, 0), (-4, -3/5), (-3, -6/5), (-2, -9/5), (-1, -12/5), (0, -3), (1, -18/5), (2, -21/5), (3, -24/5), (4, -27/5), (5, -6)}

Show that lines 4x - 3y + 1 = 0 and 4x - 3y - 3 = 0 are parallel.

• points on the line 4x - 3y + 1 = 0: {(-5, -19/3), (-4, -5), (-3, -11/3), (-2, -7/3), (-1, -1), (0, 1/3), (-1/4, 0), (1, 5/3), (2, 3), (3, 13/3), (4, 17/3), (5, 7)}
• slope of the lines: m = 4/3
• angle of inclination of the lines: 53.1°
• points on the line 4x - 3y - 3 = 0: {(-5, -23/3), (-4, -19/3), (-3, -5), (-2, -11/3), (-1, -7/3), (0, -1), (3/4, 0), (1, 1/3), (2, 5/3), (3, 3), (4, 13/3), (5, 17/3)}

Algebra: Root of a Linear Equation With One Unknown (Non-Fractional

1.	4x + 5 = 7x - 10
	Answer: x = 5
2.	3x - 4 = 8x + 41
	Answer: x = -9
3.	11x + 7 = 122 - 12x
	Answer: x = 5
4.	9x + 7 = 187 - 9x
	Answer: x = 10
5.	10x + 7 = 7x - 26
	Answer: x = -11
6.	x + 3 = 7x + 21
	Answer: x = -3
7.	8x + 7 = 37 - 2x
	Answer: x = 3
8.	3x + 11 = 9x - 13
	Answer: x = 4
9.	2x + 3 = 58 - 3x
	Answer: x = 11
10.	11x + 5 = 29 + 7x
	Answer: x = 6
11.	4x - 6 = 7 - 13x
	Answer: x = 13/17
12.	x - 4 = 5 - 5x
	Answer: x = 3/2
13.	14x + 5 = 5x + 9
	Answer: x = 4/9
14.	11x - 2 = 8x - 4
	Answer: x = -2/3
15.	6x + 2 = 7 - 14x
	Answer: x = 1/4
16.	4x - 13 = 14 - 11x
	Answer: x = 9/5
17.	4x - 3 = 4 - 10x
	Answer: x = 1/2
18.	3x + 10 = 15 - 12x
	Answer: x = 1/3
19.	5x + 7 = 10 - 10x
	Answer: x = 1/5
20.	11x - 1 = 2 - 13x
	Answer: x = 1/8

Algebra: Graphing of Linear Equation Problems