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Algebra 2
More About Linear Equations
Lesson 2-4 Part 1
Goals
Goal
• To write equations of lines.
Rubric
Level 1 – Know the goals.
Level 2 – Fully understand the
goals.
Level 3 – Use the goals to
solve simple problems.
Level 4 – Use the goals to
solve more advanced problems.
Level 5 – Adapts and applies
the goals to different and more
complex problems.
Essential Question
Big Idea: Equivalence
• Why are there different forms for the equation of a
line?
Vocabulary
• Point-Slope Form
• Standard Form
Linear Equations
Recall from Lesson 2-3 that the slope-intercept form of a linear
equation is y= mx + b, where m is the slope of the line and b is
its y-intercept. And if given the slope and y-intercept, you can
write the equation of a line in slope-intercept form.
You can also write the equation of a line given any point on
the line and the slope using the point-slope form of a linear
equation.
Point-Slope Form
If you know the slope and any point on the line, you can
write an equation of the line by using the slope formula.
Point-Slope Form
y – y1 = m(x – x1)
y-coordinate
x-coordinate
slope
(x1, y1) represents a
specific point and
(x, y) represents any
point.
Example: Writing Linear
Equations in Point-Slope Form
Write an equation in point-slope form for the line with the given slope
that contains the given point.
A.
B.
C.
y  3  4 x
y  4  x 1
Your Turn:
Write an equation in point-slope form for the line with the given slope
that contains the given point.
a.
b.
slope = 0; (3, –4)
y – (–4) = 0(x – 3)
y + 4 = 0(x – 3)
y40
Example: Writing Linear
Equations Using Point-Slope Form
Write an equation in slope-intercept form for the line with slope 3
that contains (–1, 4).
Step 1 Write the equation in point-slope form:
y – y1 = m(x – x1)
y – 4 = 3[x – (–1)]
Step 2 Rewrite the equation in slope-intercept form by solving for y.
y – 4 = 3(x + 1)
y – 4 = 3x + 3
+4
+4
y = 3x + 7
Rewrite subtraction of negative
numbers as addition.
Distribute 3 on the right side.
Add 4 to both sides.
Your Turn:
Write an equation in slope-intercept form for the line with slope
that contains (–3, 1).
Step 1 Write the equation in point-slope form:
y – y1 = m(x – x1)
Continued
Step 2 Rewrite the equation in slope-intercept form by
solving for y.
Rewrite subtraction of negative
numbers as addition.
Distribute
+1
+1
on the right side.
Add 1 to both sides.
Your Turn;
Write the equation of the line in slope-intercept
form with slope –5 through (1, 3).
Point-Slope Form
y – y1 = m(x – x1)
y – (3) = –5(x – 1)
y – 3 = –5(x – 1)
Substitute.
Simplify.
Rewrite in slope-intercept form.
y – 3 = –5(x – 1)
y – 3 = –5x + 5
y = –5x + 8
Distribute.
Solve for y.
The equation of the
line is y = –5x + 8.
Point-Slope Form
Point-Slope Form can also be used to write
the equation of a line given two points.
•
Procedure
1) Use the two given points to calculate the slope.
2) Substitute the slope and one of the given points
into the point-slope form.
3) Rewrite equation in slope-intercept form if
required.
Example: Write Equation of
Line Given Two Points
Write an equation in slope-intercept form for the line through the
two points.
(2, –3) and (4, 1)
Step 1 Find the slope.
Step 2 Substitute the slope and one of the points into the pointslope form.
y – y1 = m(x – x1)
y – (–3) = 2(x – 2)
Choose (2, –3).
Example: Continued
y – (–3) = 2(x – 2)
Step 3 Write the equation in slope-intercept form.
y + 3 = 2(x – 2)
y + 3 = 2x – 4
–3
–3
y = 2x – 7
Example: Write Equation of
Line Given Two Points
Write an equation in slope-intercept form for the line through the
two points.
(0, 1) and (–2, 9)
Step 1 Find the slope.
Step 2 Substitute the slope and one of the points into the pointslope form.
y – y1 = m(x – x1)
y – 1 = –4(x – 0)
Choose (0, 1).
Example: Continued
y – 1 = –4(x – 0)
Step 3 Write the equation in slope-intercept form.
y – 1 = –4(x – 0)
y – 1 = –4x
+ 1 +1
y = –4x + 1
Your Turn:
Write an equation in slope-intercept form for the line through the
two points.
(1, –2) and (3, 10)
Step 1 Find the slope.
Step 2 Substitute the slope and one of the points into the pointslope form.
y – y1 = m(x – x1)
y – (–2) = 6(x – 1)
y + 2 = 6(x – 1)
Choose (1, –2).
Continued
y + 2 = 6(x – 1)
Step 3 Write the equation in slope-intercept form.
y + 2 = 6(x – 1)
y + 2 = 6x – 6
–2
–2
y = 6x – 8
Your Turn:
Write an equation in slope-intercept form for the line through the
two points.
(6, 3) and (0, –1)
Step 1 Find the slope.
Step 2 Substitute the slope and one of the points into the pointslope form.
y – y1 = m(x – x1)
Choose (6, 3).
Continued
Step 3 Write the equation in slope-intercept form.
+3
+3
Your Turn:
In slope-intercept form, write the equation of the line that
contains the points in the table.
x
–8
–4
y
–5
–3.5
4
–0.5
8
1
First, find the slope. Let (x1, y1) be (–8, –5) and (x2, y2) be
(8, 1).
Next, choose a point, and use either form of the equation of a
line.
Continued
Point-Slope Form
Using (8, 1):
Rewrite in slope-intercept
form.
y – y1 = m(x – x1)
Distribute.
Substitute.
Simplify.
Solve for y.
Standard Form
Another way to determine whether a function is linear is to look
at its equation. A function is linear if it is described by a linear
equation. A linear equation is any equation that can be written in
the standard form shown below.
Standard Form
Notice that when a linear equation is written in standard
form.
Ax  By C
•x
and y are both on the same side of the equal sign.
•x
and y both have exponents of 1.
•x
and y are not multiplied together.
• x and y do not appear in denominators, exponents, or radical
signs.
Most linear equations in Algebra are first written in
slope-intercept form: y = mx + b, or point-slope
form: y – y1 = m(x – x1).
To write a linear equation in standard form you have to transform
FROM slope intercept or point slope form TO standard form.
Standard form is Ax + By = C.
A and B are not BOTH zero. A, B, and C are ALL integers
and A is positive.
In STANDARD form, the variables are on the left. The constant term is
on the right. That just means that you will have to add or subtract to get
the variables on the LEFT, the constant on the RIGHT.
Be careful if you have coefficients or the constant term that are NOT an
integer. You WILL have to get rid of the fraction (by multiplying by the
LCD).
Example: Write in standard form
2
y  x3
5
2
5  y   5( x  3)
5
5 y  2 x  15
5 y  2 x  15
2 x  5 y  15
Original Equation
Get rid of the fraction
Distribute
Subtract 2x from each side
Rearrange the terms and multiply
by -1 to make A positive.
Your Turn: Write
in standard form.
3
y  4  ( x  2)
4
3
y  4  ( x  2)
4
Given
4(y + 4) = 3(x – 2)
Multiply by 4 to get rid of the fraction.
4y + 16 = 3x – 6)
Distributive property
4y = 3x – 22
Subtract 16 from both sides
4y – 3x= – 22
Subtract 3x from both sides
– 3x + 4y = – 22
Format x before y
3x – 4y = 22
Multiply by -1 in order to get a positive
coefficient for x.
Example: Write the standard
form. Line passes through (-4,3)
with slope -2.
y  y1  m( x  x1 )
Point-slope form
y  3  2( x  (4))
Substitute
y  3  2( x  4)
Simplify
y  3  2 x  8
Distribute
2 x  y  5
Put in standard form
Your Turn: Write the standard form of
an equation of the line passing through
(-6, -3) and slope -1/2.
1
y 3   ( x  6)
2
2(y +3) = -1(x +6)
2y + 6 = -x – 6
2y = -x – 12
2y + x= -12
x + 2y = -12
Given
Multiply by 2 to get rid of the fraction.
Distributive property
Subtract 6 from both sides
Add x to both sides
Format x before y
Your Turn: Write the standard form of
an equation of the line passing through
(5, 4) and slope -2/3
2
y  4   ( x  5)
3
3(y - 4) = -2(x – 5)
Given
Multiply by 3 to get rid of the fraction.
3y – 12 = -2x +10
Distributive property
3y = -2x +22
Add 12 to both sides
3y + 2x= 22
Add 2x to both sides
2x + 3y = 22
Format x before y
Example: Write the standard form of an
equation of the line passing through (5, 4)
and (6, 3).
y2  y1
m
First find slope of the line.
x2  x1
m
3  4 1

 1 Substitute values and solve for m.
65 1
y  4  1( x  5)
y – 4 = -x + 5
y = -x + 9
y+x=9
Put into point-slope form for conversion
into Standard Form Ax + By = C
Distributive property
Add 4 to both sides.
Add x to both sides
x + y = 9 Standard form requires x come before y.
Your Turn: Write the standard form of
an equation of the line passing through
(-5, 1) and (6, -2)
y2  y1
m
First find slope of the line.
x2  x1
m
 2 1
3 3


6  (5) 6  5 11
3
( x  5)
11
11(y – 1) = -3(x + 5)
y 1  
11y – 11 = -3x – 15
11y = -3x – 4
11y + 3x = -4
3x + 11y = -4
Substitute values and solve for m.
Put into point-slope form for conversion
into Standard Form Ax + By = C
Multiply by 11 to get rid of fraction
Distributive property
Add 4 to both sides.
Add x to both sides
Standard form requires x come before y.
Caution
When writing a linear equation, NEVER leave it
in point-slope form, always transform it to either
slope-intercept form or standard form.
Linear Equations
Essential Question
Big Idea: Equivalence
• Why are there different forms for the equation of a
line?
• Depending on what information you know about
the line, use the form that is easiest to write an
equation of that line.
Assignment
• Section 2-4 Part 1, Pg 91; #1 – 6 all, 8 – 22
even.