LINEAR EQUATIONS MSJC ~ Menifee Valley Campus Math Center Workshop Series Janice Levasseur.

LINEAR EQUATIONS

MSJC ~ Menifee Valley Campus

Math Center Workshop SeriesJanice Levasseur

Equations of the Line

• Write the equation of a line given the slope and the y-intercept

• Write the equation of a line given the slope and a point

• Write the equation of a line given two points


Write the equation of a line given the slope and the y-intercept: m and (0, b)



Ex: Find an equation of the line with slope = 6 and y-int = (0, -3/2)

Recall: slope-intercept form of a linear equation

y = mx + b, where m and b are constants

Given the y-int = (0, -3/2) b = - 3/2

Given the slope = 6 m = 6

Putting everything together we get the equation of the line in slope-int form:

y = m x + b6 - 3/2

y = 6x – 3/2

Ex: Find an equation of the line with slope = 1.23 and y-int = (0, 0.63)

Recall: slope-intercept form of a linear equation

y = mx + b, where m and b are constants

Given the y-int = (0, 0.63) b = 0.63

Given the slope = 1.23 m = 1.23

Putting everything together we get the equation of the line in slope-int form:

y = m x + b1.23 0.63

y = 1.23x + 0.63



Write the equation of a line given the slope and a point: m and (x1, y1)


Ex: Find an equation of the line with slope = -3 that contains the

point (4, 2) Start with the slope-intercept form of a linear equation y = mx + b

Slope = - 3 y = - 3x + b What is b, though?What is b, though?

Use the given point (4, 2) x = 4 and y = 2

y = - 3x + b 2 = - 3(4) + b

2 = -12 + b

14 = b

put it together

we have m and b

y = - 3 x + 14

Ex: Find an equation of the line with slope = -0.25 that contains

the point (2, -6) Start with the slope-intercept form of a linear equation y = mx + b

Slope = -0.25 y = -0.25 x + b What is b, though?

Use the given point (2, -6) x = 2 and y = -6

y = -0.25 x + b -6 = -0.25(2) + b

-6 = -0.5 + b

-5.5 = b

put it together

we have m and b

y = -0.25x – 5.5




Write the equation of a line given two points: (x1, y1) and (x2, y2)

Ex: Find an equation of the line containing the points (-2, 1) and (3, 5)

First, find the slope of the line containing the points:

Slope = m = rise = y1 - y2 = 1 – (5)

run x1 - x2 -2 – 3

Point 1 Point 2

54

54

Now we have m = 4/5 and two points. Pick one point and proceed like in the last section.

We have m = 4/5, the point (-2, 1), and y = mx + b

Slope = 4/5 y = 4/5x + b What is b, though?

Use the given point (-2, 1) x = -2 and y = 1

y = 4/5x + b 1 = 4/5(-2) + b

1 = (-8/5) + b

13/5 = b

put it together we have m and b

y = 4/5x + 13/5

Ex: Find an equation of the line containing the points (-4, 5) and (-2, -3)


Slope = m = rise = y1 - y2 = 5 – (-3)

run x1 - x2 -4 – (-2)

Point 1 Point 2

28

Now we have m = -4 and two points. Pick one point and proceed like in the last section.

= -4

We have m = -4, the point (-4, 5), and y = mx + b

Slope = -4 y = -4x + b What is b, though?

Use the given point (-4, 5) x = -4 and y = 5

y = -4x + b 5 = -4(-4) + b

5 = 16 + b

-11 = b


y = -4x – 11

Ex: Find an equation of the line containing the points (0, 0) and (1, -5)


Slope = m = rise = y1 - y2 = 0 – (-5)

run x1 - x2 0 – (1)

Point 1 Point 2

15

Now we have m = -5 and two points. Pick one point and proceed like in the last section.

= -5

We have m = -5, the point (0, 0), and y = mx + b

Slope = -5 y = -5x + b What is b, though?

Use the given point (0, 0) x = 0 and y = 0

y = -5x + b 0 = -5(0) + b

0 = 0 + b

0 = b


y = -5x + 0

y = -5x

Parallel & Perpendicular Lines

• When we graph a pair of linear equations, there are three possibilities:

1. the graphs intersect at exactly one point

2. the graphs do not intersect

3. the graphs intersect at infinitely many points

• We will consider a special case of situation 1 and also situation 2.

Perpendicular Lines (Situation 1)

• Perpendicular lines intersect at a right angle

• Notation: › L1: y = m1x + b1

› L2: y = m2x + b2

› L1 L2

Nonvertical perpendicular lines have slopes that are the negative reciprocals of each other:

m1m2 = -1 ~ or ~ m1 = - 1/m2 ~ or ~ m2 = - 1/m1

If l1 is vertical (l1: x = a) and is perpendicular to l2, then l2 is horizontal (l2: y = b) ~ and ~ vice versa

Ex: Determine whether or not the graphs of the equations of the lines

are perpendicular:l1: x + y = 8 and l2: x – y = - 1

First, determine the slopes of each line by rewriting the equations in slope-intercept form:

l1: y = - x + 8 and l2: y = x + 1

m1 = and m2 =

1 1

-1 1

Since m1m2 = (-1)(1) = -1, the lines are perpendicular.


are perpendicular:l1: -2x + 3y = -21 and l2: 2y – 3x = 16

First, determine the slopes of each line by rewriting the equations in slope-intercept form:

l1: y = (2/3)x - 7 and l2: y = (3/2)x + 8

m1 = and m2 = 2/3 3/2

Since m1m2 = (2/3)(3/2) = 1 = -1

Therefore, the lines are not perpendicular!

Parallel Lines (Situation 2)

• Parallel lines do not intersect

• Notation: › L1: y = m1x + b1

› L2: y = m2x + b2

› L1 L2

Nonvertical parallel lines have the same slopes but different y-intercepts:

m1 = m2 ~ and ~ b1 = b2

Horizontal Parallel Lines have equations

y = p and y = q where p and q differ.

Vertical Parallel Lines have equations

x = p and x = q where p and q differ.


are parallel:l1: 3x - y = -5 and l2: y – 3x = - 2

First, determine the slopes and intercepts of each line by rewriting the equations in slope-intercept form:

l1: y = 3x + 5 and l2: y = 3x - 2

m1 = and m2 = 3 3

Since m1 = m2 and b1 = b2 the lines are parallel.

b1 = and b2 = 5 -2


are parallel:l1: 4x + y = 3 and l2: x + 4y = - 4

First, determine the slopes and intercepts of each line by rewriting the equations in slope-intercept form:

l1: y = -4x + 3 and l2: y = (-¼)x - 1

m1 = and m2 = -4 - ¼

Since m1 = m2 the lines are not parallel.

LINEAR EQUATIONS MSJC ~ Menifee Valley Campus Math Center Workshop Series Janice Levasseur.

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