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Forms of Linear Equations
Slope-Intercept Form
y = mx + b
Algebra 1
Section 5.6
Point-Slope Form
of the Equation of a Line
Standard Form
Ax + By = C
Point-Slope Form
y – y1 = m(x – x1)
when given a point and a slope, this is
another way to get the equation of the line
Write the equation of the line that passes through the point and has
the given slope. Then rewrite the equation in slope-intercept form.
(x1,y1)
1. (2,4), m = 3
y – y1 = m(x – x1) (point-slope form)
y – 4 = 3(x – 2) substitute in for x1 and y1
y – 4 = 3x – 6 distributive property (get in slope-intercept form)
+4
+4
y = 3x – 2
Mr. Sims
Write the equation of the line that passes through the point and has
the given slope. Then rewrite the equation in slope-intercept form.
(x1,y1)
2. (-6,-2), m = ½
y – y1 = m(x – x1) (point-slope form)
y – (-2) = ½[x –(-6)] substitute in for x1and y1
y + 2 = ½(x + 6)
y + 2 = ½x + 3 distributive property to get into slope-intercept form
-2
-2
y = ½x + 1
(x1,y1)
3. (0,-3), m = ¾
y – y1 = m(x – x1) (point-slope form)
y – (-3) = ¾(x – 0) substitute
y + 3 = ¾x distributive property
-3
-3
y = ¾x – 3
slope-intercept form
Mr. Sims
Write the equation of the line that passes through the point and has
the given slope. Then rewrite the equation in standard form.
4. (7,-10), m = -6
y – y1 = m(x – x1) (point-slope form)
y – (-10) = -6(x – 7) substitute
y + 10 = -6x + 42 distributive property to get in standard form
+6x -10
+6x -10
6x + y = 32
standard form
5. (-8,-12), m = ¼
y – y1 = m(x – x1) (point-slope form)
y – (-12) = ¼[x – (-8)] substitute
y + 12 = ¼(x + 8)
y + 12 = ¼x + 2 distributive property
-¼x -12 -¼x -12
- 4 -¼x + y = -10
multiply by – 4 to get rid of fraction
x – 4y = 40 (standard form)
Mr. Sims
Write the equation of the line that passes through the two points.
Final equation should be in slope-intercept form.
6. (0,2) , (3,-1)
m
y y
2
1
2
1
x x
1  2

30
3

3
find slope
choose either point and use the slope
to substitute into point-slope form
(0,2), m = -1
y – y1 = m(x – x1) (point-slope form)
y – 2 = -1(x – 0) substitute
y – 2 = -x distributive property
+2
+2
y = -x + 2
slope-intercept form
m = -1
Mr. Sims
Write the equation of the line that passes through the two points.
Final equation should be in slope-intercept form.
7. (-3,4) , (8,4)
m
y y
2
1
2
1
x x
44

8  (3)
find slope
(8,4), m = 0
y – y1 = m(x – x1)
y – 4 = 0(x – 8) substitute
y – 4 = 0 substitute
+4 +4
y=4
slope-intercept form
0

11
m=0
Mr. Sims
Write the equation of the line that passes through the two points.
Final equation should be in standard form.
8. (- 4,0) , (2,-2)
m
y y
2
1 find slope
2
1
x x
20

2  (4)
2

6
1
m
3
1
(4,0), m  
3
y – y1 = m(x – x1)
(point-slope form)
1
y  0   [x  (4)] substitute
3
1
multiply by –3 to
y


[x

4]
-3
get rid of fraction
3
(standard form)
-3y = x + 4
-x
-x
-x – 3y = 4
standard form
Mr. Sims
Write the equation of the line that passes through the two points.
Final equation should be in standard form.
9. (6,0) , (0,-3)
m
y y
2
1
2
1
x x
30

06

3
6
1
m
2
find slope
1
(6,0), m 
2
y – y1 = m(x – x1)
(point-slope form)
1
y  0  (x  6) substitute
2
1
2 y  x  3 distributive property
2
2y = x – 6
-x
multiply by 2 to get rid of
fraction
-x
-x + 2y = -6
Mr. Sims
total cost = entrance fee + $2(# of rides)
y
=
3
+
2x
Mr. Sims
(x)(7) = 322
x  322
7
divide both sides by 7
x = 46
Mr. Sims
If two angles of a triangle are congruent, then
the sides opposite them are congruent
YZ  XY
Mr. Sims
Write an equation of the line that passes through the
point and has the given slope in slope-intercept form.
1. ) (2,4) , m = -2
3.)
(3,2) , m  
Algebra 1
Section 5.6 A
Assignment
2.) (-2,3) , m = 2
1
2
5.)
4.)
Mr. Sims
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