# 2.5 Writing Linear Equations

```2.5 Writing Linear
Equations
Algebra 2
Mrs. Spitz
Fall 2006
Objectives:
 Write the slope-intercept form of an equation
given the slope and a point or two points,
 Write the standard form of an equation given
the slope and a point, or two points and
 Write an equation of a line that is parallel or
perpendicular to the graph of a given
equation.
Assignment
 pp. 76-78 #6-41
Introduction
 In Lesson 2.2, you learned if a function can
be written in the form y = mx + b, then it is a
linear equation. But what numbers do m and
b represent?
Finding Slope-Intercept form
 Look at the graph. The
line passes through
points A(0, b) and C(x,
y). Notice that b is the y
intercept of AC.
Suppose you need to
find the slope of AC.
 Now solve the equation
for y.
y b
m
x0
y b
m
x
Finding Slope-Intercept form
y b
m
x
mx y b
m x b  y
y  m x b
Multiply each side by x.
Symmetric property of Equality
You may recognize this as the form of a linear function. When an
equation is written in the form y = mx + b, it is in slope-intercept form.
Slope Intercept Form of a Linear
Equation
The slope-intercept form of the equation of a
line is y = mx + b, where m is the slope and b
is the y-intercept.
Notes
 If you are given the slope
and y-intercept of a line, you
can find an equation of the
line by substituting values of
m and b into the slopeintercept form of the
equation. Then the equation
can be written in standard
form. For example, if you
know the slope of a line is ⅔
and the y-intercept is 5, an
equation of the line is
2
y  x6
3
In standard form, 2x – 3y = -18
Ex. 1: Find the slope-intercept form of the equation of
the line that has a slope of ¾ and passes through (8, 2).
 You know the slope and
the x and y values of
one point on the graph.
Substitute for m, x and
y in the slope-intercept
form.
y  m x b
3
2  ( )(8)  b
4
2  6b
4b
Slope-Intercept form
The equation in slopeintercept form is
Substitute values
y = ¾x - 4
Multiply
Subtract 6 from both sides
Standard Form
 Remember that the
stand form of the
equation of a line is Ax
+ By = C. Suppose we
write this general
equation in slopeintercept form.
Ax  By  C
By   Ax  C
A
C
y  x
B
B
The slope is
y-intercept is
A and the
B
C
, for B ≠ 0.
B

This can be used to write an
equation in standard form when
you are given the information you
usually use to find the slopeintercept form.
Subtract Ax from each side.
Divide each side by B.
Ex. 2: Find the standard form of the equation that passes
through (-2, 5) and (3, 1).
 First use the two given
points to find the slope
of the line.
m
y 2  y1
x2  x1
1 5
3  ( 2)
4
m
5
m
If the slope is -4/5, then –A/B = (4/5). Thus A = 4 and B = 5
Substitute these values into the
standard form. The resulting
equation is 4x + 5y = C. Since
one of the points on the line is
(3, 1), you can substitute these
values into the equation to find
C.
Ex. 2: Find the standard form of the equation that passes
through (-2, 5) and (3, 1).
Ax  By  C
Standard form of a linear equation.
4x  5 y  C
Substitute values for A and B.
4(3)  5(1)  C
Substitute values for x and y.
12  5  C
Simplify
17  C
Ex. 3: Application problem
 The atmospheric
pressure at sea level is
14.7 pounds per square
inch. As divers go
deeper into the ocean,
the pressure increases.
Use the chart to write
an equation in slopeintercept form that
approximates this
relationship. Then find
the pressure at 30,000
feet below sea level.
Depth
Pressure
(in feet)
(lbs/in2)
Sea level (0)
14.7
600
269
1200
536
3000
1338
7200
3208
18,000
8019
How?
 Let x represent the ocean dept and y
represent the pressure.
 Use your calculator and a pair of points to
find the slope of the line.
269  536
m
 0.445
600  1200
 You may want to use another set of points to
3208  1338
m
 0.445238095
7200  3000
Next?
 The slope of the line is approximately 0.445. The y-
intercept corresponds to the ocean depth of 0 (at sea
level). So the y-intercept is 14.7. An equation that
approximates the pressure at certain ocean depths is
y= 0.445x + 14.7
 The equation that is derived may differ based on the
set of points used to determine the slope.
 Use your calculator again to approximate pressure at
30,000.
0.445 x 30,000 + 14.7 = 13364.7
The result is 13,364.7 pounds per square inch.
Ex. 4: Write an equation of the line that passes through
(4, 6) and is parallel to the line whose equation is: y =
2/3x + 5.
 Parallel lines
have the same
slope, so the
slope of both
lines is 2/3.
Use the slopeintercept form
and the point
(4, 6) to find
the equation.
y  m x b
2
6  ( )(4)  b
3
8
6  b
3
18 8
 b
3 3
10
b
3
Slope-Intercept form
Substitute values for x, y and
m
Multiply 2/3 by 4
Common denominator
Subtract 8/3 from 18/3 to get
the y-intercept.
The y-intercept is 10/3. An equation of the line is
y
2
10
x
3
3
What’s the standard form?
2
10
y  x
3
3
A2
B3
C  10
Ax  By  C
2 x  3 y  10
2
10
y  x
3
3
Ex. 5: Write an equation in standard form for the line that passes
through (4, 6) and is perpendicular to the line whose equation is
y=2/3x + 5.
3
m
2
A3
B2
The slope of the given line is 2/3. Since
the product of this slope and the slope
of a perpendicular line is -1, the slope of
a perpendicular line is
-3/2. You can use the slope -3/2 and the
point (4,6) to write the equation in
standard form.
Ax  By  C
Standard Form
3(4)  2(6)  C
Substitute values for A, B and x
and y
12  12  C
24  C
3x  2 y  24
Multiply values
Substitute into the standard form.
Ex. 2: Find the standard form of the equation that passes
through (-2, 5) and (3, 1).
 First use the two given
points to find the slope
of the line.
y  m x b
3
2  ( )(8)  b
4
2  6b
4b
Slope-Intercept form
The equation in slopeintercept form is
Substitute values
y = ¾x - 4
Multiply
Subtract 6 from both sides
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