1. The document provides examples of writing linear equations in slope-intercept form, point-slope form, and standard form. It also includes examples of finding the slope and y-intercept of a line from its graph and writing equations of lines given certain points on the line. 2. There are guided practice problems that have students write equations of lines passing through given points, with one point and the slope given, and finding missing coefficients in standard form equations. 3. The examples and guided practice cover skills related to writing, manipulating, and identifying components of linear equations.

1. Rewrite the following equations in
Column A in the form 𝑦 = 𝑚𝑥 + 𝑏,
and find its corresponding slope-
intercept form. (2 pts.each)
LEARNING TASK 1

2. COLUMN A
_1. 𝑥 + 𝑦 = 3
_2. 𝑥 − 3𝑦 = 9
_3. 10𝑥 − 2𝑦 = 20
_4. 6𝑥 = 12 + 2𝑦
_5. 4𝑥 + 16 − 8𝑦 = 0

3. Write an equation of the line in STANDARD
FORM using the information given.
5.) m = 2 and (3,-2)
Start with Point-Slope Form

y  y1  m(x  x1)

y  2  2(x  3)
Now put into slope-intercept form

y  2  2x  6
-2 -2

y  2x  8
-2x -2x
2x  3y 12
Now put into Standard form
No LEADING NEGATIVES!
Change all the signs of each term
2x  3y  12

4. ANSWER y + 4 = 2(x – 6)
Daily Homework Quiz Review 5.3
Write an equation in point-slope form of the line
that passes through (6, –4) and has slope 2.
1.
Write an equation in point-slope form of the line
that passes through (–1, –6) and (3, 10).
2.
ANSWER y + 6 = 4(x + 1) or y –10 = 4(x–3)
y – y1 = m(x – x1)
m 
10  6
3  1

16
4
 4

5. Review 5.3
ANSWER
ANSWER
1. (1, 4), (6, –1)
y + 2 = 3(x + 1) or y – 7 = 3(x – 2)
y – 4 = –1(x – 1) or y + 1 = –1(x – 6)
2. (–1, –2), (2, 7)
Write an equation in point-slope form of the line that
passes through the given points.
NOW PUT EACH ANSWER IN SLOPE-INTERCEPT FORM!!
y – 4 = – x + 1 or y + 1 = - x + 6
y = – x + 5 or y = - x + 5
+ 4 + 4 - 1 - 1
y + 2 = 3x + 3 or y – 7 = 3x – 6
- 2 - 2 + 7 + 7
y = 3x + 1 or y = 3x + 1
m 
1 4
6 1

5
5
 1

6. 5.4 Write Linear Equations in
Standard Form

7. Convert this equation into
standard form:
1.)

y 
2
5
x  3 Multiply everything by 5

5y  2x 15
-2x -2x
Ax + By = C

2x  5y  15
Move over the “x” term
I CAN’T LEAD WITH A NEGATIVE!!!
So let’s change the sign of each term.
2x  5y 15
It’s kind of like
moving backwards!

8. Convert this equation into
standard form:
2.)

y  x  5
+ x + x
Ax + By = C
Move over the “x” term

x  y  5

9. Convert this equation into
standard form:
3.)

y 
1
2
x  7 Multiply everything by 2

2y  1x 14
+ 1x + 1x
Ax + By = C
Move over the “x” term

x  2y 14

10. Convert this equation into
standard form:
4.)

y 
2
3
x  4 Multiply everything by 3

3y  2x 12
-2x -2x
Ax + By = C

2x  3y 12
Move over the “x” term
I CAN’T LEAD WITH A NEGATIVE!!!
So let’s change the sign of each term.
2x  3y  12

11. Write an equation of the line in STANDARD
FORM using the information given.
5.) m = and (4,-5)
Start with Point-Slope Form

y  y1  m(x  x1)

y  5 
3
2
(x  4)
Now put into slope-intercept form

y  5 
3
2
x  6
- 5 - 5
y 
3
2
x 11
- 3x - 3x
3x  2y  22
Now put into Standard form
No LEADING NEGATIVES!
Change all the signs of each term 3x  2y  22

3
2
Multiply everything by 2
2y  3x  22

12. Every one get communicators with a blank
side!!!

13. Write an equation of the line in STANDARD
FORM using the information given.
5.) (-4,4) and (0,3)
Start with Point-Slope Form

y  y1  m(x  x1)

y  4 
1
4
(x  4)
Now put into slope-intercept form

y  4 
1
4
x 1
+ 4 + 4
y 
1
4
x  3
+1x +1x
x  4y 12
Now put into Standard form
No LEADING NEGATIVES!
Change all the signs of each term
Multiply everything by 4
4y  1x 12

m 
3  4
0  4

1
4

14. Write the point-slope form of the line
that passes through (4,3) and (1,2)

15. Write the slope-intercept form of the
line that passes through (4,5) and (1,-1)

16. Write an equation of the line in STANDARD
FORM using the information given.
5.) m = -2 and (-4,3)
Start with Point-Slope Form

y  y1  m(x  x1)

y  3  2(x  4)
Now put into slope-intercept form

y  3  2x  8
+ 3 + 3

y  2x  5
+ 2x + 2x
2x  y  5
Now put into Standard form

17. Write an equation of the line in STANDARD
FORM using the information given.
5.) m = -3 and (3,-5)
Start with Point-Slope Form

y  y1  m(x  x1)

y  5 
3
2
(x  4)
Now put into slope-intercept form

y  5 
3
2
x  6
- 5 - 5
y 
3
2
x 11
- 3x - 3x
3x  2y  22
Now put into Standard form
No LEADING NEGATIVES!
Change all the signs of each term 3x  2y  22
Multiply everything by 2
2y  3x  22

18. Write an equation of the line in STANDARD
FORM using the information given.
5.) (4,0) and (0,3)
Start with Point-Slope Form

y  y1  m(x  x1)

y  0 
3
4
(x  4)
Now put into slope-intercept form

y 
3
4
x  3
+ 3x + 3x
Now put into Standard form
3x  4y 12
Multiply everything by 4
4y  3x 12

m 
3  0
0  4

3
4

3
4

19. Write an equation of the line in STANDARD
FORM using the information given.
5.) (2,0) and (0,5)
Start with Point-Slope Form

y  y1  m(x  x1)

y  0 
5
2
(x  2)
Now put into slope-intercept form

y 
5
2
x  5
+ 5x + 5x
Now put into Standard form
5x  2y 10
Multiply everything by 2
2y  5x 10

m 
5  0
0  2

5
2

5
2

20. SOLUTION
y – y1 = m(x – x1)
Calculate the slope.
STEP 1
EXAMPLE 2 Write an equation from a graph
–3
m =
1 – (–2)
1 – 2
=
3
–1 =
Write an equation in point-slope form. Use (1, 1).
Write point-slope form.
y – 1 = –3(x – 1) Substitute 1 for y1, 3 for m
and 1 for x1.
Write an equation in standard form of the line shown.
STEP 2

21. Rewrite the equation in standard form.
EXAMPLE 2 Write an equation from a graph
3x + y = 4 Simplify. Collect variable
terms on one side,
constants on the other.
STEP 3

22. EXAMPLE 2 Write an equation from a graph
GUIDED PRACTICE for Examples 1 and 2
Write an equation in standard form of the line
through (3, –1) and (2, –3).
2.
–2x + y = –7
ANSWER

23. Simplify.
Find the value of A. Substitute the
coordinates of the given point for x and y in
the equation. Solve for A.
STEP 1
SOLUTION
EXAMPLE 4
Find the missing coefficient in the equation of the line
shown. Write the completed equation.
Ax + 3y = 2
A(–1) + 3(0) = 2
–A = 2
A = –2
Write equation.
Substitute –1 for x and 0 for y.
Divide by –1.
EXAMPLE 3
EXAMPLE 4 Complete an equation in standard form

24. EXAMPLE 4 Complete an equation in standard form
Complete the equation.
–2x + 3y = 2 Substitute –2 for A.
STEP 2

25. Write equations of the horizontal and vertical lines that
pass through the given point.
GUIDED PRACTICE for Examples 3 and 4
3. (–8, –9)
y = –9, x = –8
ANSWER

26. GUIDED PRACTICE for Examples 3 and 4
4. (13, –5)
y = –5, x = 13
ANSWER
Write equations of the horizontal and vertical lines that
pass through the given point.

27. EXAMPLE 4 Complete an equation in standard form
Find the missing coefficient in the equation of the
line that passes through the given point. Write the
completed equation.
EXAMPLE 3 Write an equation of a line
GUIDED PRACTICE for Examples 3 and 4
5. –4x + By = 7, (–1, 1)
ANSWER 3; –4x + 3y = 7

28. To write another equivalent
equation, multiply each side
by 0.5.
4x – 12y = 8
To write one equivalent
equation, multiply each
side by 2.
SOLUTION
Write two equations in standard form that are equivalent
to 2x – 6y = 4.
EXAMPLE 1 Write equivalent equations in standard form
x – 3y = 2

29. EXAMPLE 1
GUIDED PRACTICE for Examples 1 and 2
Write two equations in standard form that are
equivalent to x – y = 3.
1.
2x – 2y = 6, 3x – 3y = 9
ANSWER

30. Substitute 0 for s.
8(0) + 12l = 144
l = 12
Substitute 0 for l.
s = 18
8s + 12(0) = 144
ANSWER
The equation 8s + 12l = 144 models the possible
combinations.
b. Find the intercepts of the graph.
EXAMPLE 5 Solve a multi-step problem

31. EXAMPLE 4 Complete an equation in standard form
EXAMPLE 3 Write an equation of a line
GUIDED PRACTICE for Examples 3 and 4
6. Ax + y = –3, (2, 11)
Find the missing coefficient in the equation of the
line that passes through the given point. Write the
completed equation.
ANSWER –7; –7x +y = –3