5.4 Standard form
• Obj: I can write an equation of a line in
standard form.
• Write down new HW.
• Ch 5 Retests are available in WIN.(offer expires
Friday and Im out tomorrow)
• Warm up: Convert from point slope form to
slope intercept form
y-5 = -3(x-10)
• Find the x and y intercepts to -4x + 8y =16
5.4 Write Linear Equations in
Standard Form


Convert this equation into
standard form: Ax + By = C
2
y  x3
5
1.)
Multiply everything by 5

5y  2x 15
-2x
Move over the “x” term
-2x
2x  5y  15
2x  5y  15
I CAN’T LEAD WITH A NEGATIVE!!!
So let’s change the sign of each term.
It’s kind of like
moving backwards!
Convert this equation into
standard form: Ax + By = C
2.)
y  x  5
+x
+x

xy 5

Move over the “x” term
Convert this equation into
standard form: Ax + By = C
3.)
1
y
x7
2
Multiply everything by 2
2y  1x 14
Move over the “x” term

+ 1x


+ 1x
x  2y  14
Convert this equation into
standard form: Ax + By = C
2
y  x4
3
Multiply everything by 3
3y  2x 12
Move over the “x” term
4.)

-2x


-2x
2x  3y  12
2x  3y  12
I CAN’T LEAD WITH A NEGATIVE!!!
So let’s change the sign of each term.
Write an equation of the line in STANDARD
FORM using the information given.
5.) m = 2 and (3,-2)
y  y1  m(x  x1)
Start with Point-Slope Form
y  2  2(x  3)
Now put into slope-intercept form

y  2  2x  6
-2

Now put into Standard form
y  2x  8
-2x

-2
-2x
 2 x  y  12
No LEADING NEGATIVES!
Change all the signs of each term

2x  y  8
Write an equation of the line in STANDARD
FORM using the information given.
3
5.) m =
and (4,-5)
2
y  y1  m(x  x1)
3
y  5  (x  4)
2
put into slope-intercept form
Now
3
y 5  x 6

2 -5
-5
Now put into Standard form
3
y  x 11
2y  3x  22

2
Multiply everything by 2
- 3x
- 3x
Start with Point-Slope Form

No LEADING NEGATIVES!
Change all the signs of each term
3x  2y  22
3x  2y  22
Write an equation of the line in STANDARD
FORM using the information given.
m
34
1

0  4 4
5.) (-4,4) and (0,3)
y  y1  m(x  x1)
Start with Point-Slope Form

1
y  4  (x  4)
4
Now put into slope-intercept form
1
y 4 
x 1

4 +4
+4
Now put into Standard form
1
y
x3
4 y  1x 12

4
Multiply everything by 4
+1x
+1x

No LEADING NEGATIVES!
x  4 y  12
Change all the signs of each term
Write the point-slope form of the line
that passes through (4,3) and (1,2)
Write the slope-intercept form of the
line that passes through (4,5) and (1,-1)
Write an equation of the line in STANDARD
FORM using the information given.
5.) m = -2
and (-4,3)
y  y1  m(x  x1)
Start with Point-Slope Form
y  3  2(x  4)
Now put into slope-intercept form
y  3  2x  8


Now put into Standard form


+3
y  2x  5
+ 2x
+ 2x
2x  y  5
+3
Write an equation of the line in STANDARD
FORM using the information given.
5.) m = -3
and (3,-5)
y  y1  m(x  x1)
Start with Point-Slope Form
3
y  5  (x  4)
2
Now put into slope-intercept form
3
y 5  x 6

2 -5
-5
Now put into Standard form
3
y  x 11
2y  3x  22

2
Multiply everything by 2
- 3x
- 3x

No LEADING NEGATIVES!
3x  2y  22
Change all the signs of each term
3x  2y  22
Write an equation of the line in STANDARD
FORM using the information given.
5.) (4,0) and (0,3)
Start with Point-Slope Form
Now put into slope-intercept form

m
30
3 3


0  4 4 4
y  y1  m(x  x1)

3
y 0 
(x  4)
4
3
y
x3
4
Now put into Standard form
4 y  3x 12

Multiply everything by 4

+ 3x
3x  4 y  12
+ 3x
Write an equation of the line in STANDARD
FORM using the information given.
m
5.) (2,0) and (0,5)
Start with Point-Slope Form
Now put into slope-intercept form

Now put into Standard form

5 0 5
5


0  2 2 2
y  y1  m(x  x1)

5
y 0 
(x  2)
2
5
y
x 5
2
2y  5x 10
Multiply everything by 2
+ 5x

+ 5x
5x  2y  10
EXAMPLE 2
Write an equation from a graph
Write an equation in standard form of the line shown.
SOLUTION
STEP 1
Calculate the slope.
1 – (–2)
3
m=
= –1 = –3
1–2
STEP 2
Write an equation in point-slope form. Use (1, 1).
y – y1 = m(x – x1)
y – 1 = –3(x – 1)
Write point-slope form.
Substitute 1 for y1, 3 for m
and 1 for x1.
EXAMPLE 2
Write an equation from a graph
STEP 3
Rewrite the equation in standard form.
3x + y = 4
Simplify. Collect variable
terms on one side,
constants on the other.
from 1a and
graph
EXAMPLE
2 Write an equation
for Examples
2
GUIDED PRACTICE
2. Write an equation in standard form of the line
through (3, –1) and (2, –3).
ANSWER
–2x + y = –7
EXAMPLE 4
3
Complete an equation in standard form
Find the missing coefficient in the equation of the line
shown. Write the completed equation.
SOLUTION
STEP 1
Find the value of A. Substitute the
coordinates of the given point for x and y in
the equation. Solve for A.
Ax + 3y = 2
A(–1) + 3(0) = 2
–A = 2
A = –2
Write equation.
Substitute –1 for x and 0 for y.
Simplify.
Divide by –1.
EXAMPLE 4
Complete an equation in standard form
STEP 2
Complete the equation.
–2x + 3y = 2
Substitute –2 for A.
GUIDED PRACTICE
for Examples 3 and 4
Write equations of the horizontal and vertical lines that
pass through the given point.
3.
(–8, –9)
ANSWER
y = –9, x = –8
GUIDED PRACTICE
for Examples 3 and 4
Write equations of the horizontal and vertical lines that
pass through the given point.
4.
(13, –5)
ANSWER
y = –5, x = 13
an
equation
standard
form
EXAMPLE
4
3 Complete
for
Examples
3 in
and
4
Write an
equation
of
a line
GUIDED PRACTICE
Find the missing coefficient in the equation of the
line that passes through the given point. Write the
completed equation.
5.
–4x + By = 7, (–1, 1)
ANSWER
3; –4x + 3y = 7
EXAMPLE 1
Write equivalent equations in standard form
Write two equations in standard form that are equivalent
to 2x – 6y = 4.
SOLUTION
To write one equivalent
equation, multiply each
side by 2.
4x – 12y = 8
To write another equivalent
equation, multiply each side
by 0.5.
x – 3y = 2
EXAMPLE
1
GUIDED PRACTICE
for Examples 1 and 2
1. Write two equations in standard form that are
equivalent to x – y = 3.
ANSWER
2x – 2y = 6, 3x – 3y = 9
EXAMPLE 5
Solve a multi-step problem
ANSWER
The equation 8s + 12l = 144 models the possible
combinations.
b.
Find the intercepts of the graph.
Substitute 0 for s.
8(0) + 12l = 144
l = 12
Substitute 0 for l.
8s + 12(0) = 144
s = 18
an
equation
standard
form
EXAMPLE
4
3 Complete
for
Examples
3 in
and
4
Write an
equation
of
a line
GUIDED PRACTICE
Find the missing coefficient in the equation of the
line that passes through the given point. Write the
completed equation.
6.
Ax + y = –3, (2, 11)
ANSWER
–7; –7x +y = –3