PPT Chapter 5 Reviewx

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1.Given slope (m) and y-intercept (b) create the equation in slopeintercept form.
2. Look at a graph and write an equation of a line in slopeintercept form.
3. Know how to plug into point-slope form. y  y1  m(x  x1)
y  y1
m 2
4. Find the slope between two points.
x 2  x1
5. Write an equation of a line that passes through two points.

6. Find an equation of a line that is parallel to an equation given
and also given a random point.

7. Find an equation of a line given a slope and a random point.
8. Decide which two lines are parallel.
9. Convert an equation into standard form.
10. Write an equation of a horizontal (y = #) or vertical line (x = #).
11. Decide which two lines are perpendicular.
12. Look at an equation of a line and find the slope of that line.
Slope-Intercept Form
Standard Form
y  mx  b
m  slope of the line
b  y  intercept
Ax  By  C
A, B, and C are integers
A  0, A must be postive
y  y1  m  x  x1 
Point-Slope Form
m  slope of the line
 x1 , y1  is any point
Let’s try one…
Given “m” (the slope remember!) = 2
And “b” (the y-intercept) = +9
All you have to do is plug those values into
y = mx + b
The equation becomes…
y = 2x + 9
Given m = 2/3, b = -12,
Write the equation of a line in slope-intercept
form.
Y = mx + b
Y = 2/3x – 12
*************************
One last example…
Given m = -5, b = -1
Write the equation of a line in slope-intercept
form.
Y = mx + b
Y = -5x - 1
GUIDED PRACTICE
for Example 1
Write an equation of the line that has the given slope
and y-intercept.
1.
m = 3, b = 1
ANSWER
y=3x+1
2.
m = –2 , b = –4
ANSWER
y = –2x – 4
3.
m=–
3, b =
4
ANSWER
7
3
y=–
x+
4
2
7
2
1) m = 3, b = -14
y = 3x - 14
2) m = -½, b = 4
y =-½x + 4
3) m = -3, b = -7
y =-3x - 7
4) m = 1/2 , b = 0
y = ½x
5)
m = 2, b = 4
y =2x + 4
6)
m = 0, b = -3
y=-3
Write an equation given the slope and y-intercept
Write an equation of the line shown in slope-intercept form.
m=¾
b = (0,-2)
y = ¾x - 2
True
False
a)
b)
c)
d)
-4/3
-3/4
4/3
-1/3
a)
b)
c)
d)
x = -5
y=7
x=y
x+y=0
Point-Slope Form
Standard Form
Using point-slope form, write the equation of a line
that passes through (4, 1) with slope -2.
y – y1 = m(x – x1)
y – 1 = -2(x – 4)Substitute 4 for x , 1 for y and -2 for m.
1
Write in slope-intercept form.
y – 1 = -2x + 8 Add 1 to both sides
y = -2x + 9
1
Using point-slope form, write the equation of a
line that passes through (-1, 3) with slope 7.
y – y1 = m(x – x1)
y – 3 = 7[x – (-1)]
y – 3 = 7(x + 1)
Write in slope-intercept form
y – 3 = 7x + 7
y = 7x + 10
4--4
y2 – y1
=
m=
x2 – x1
-1-3
8
=
–4
y2 – y1 = m(x – x1)
Use point-slope form.
y + 4 = – 2(x – 3)
Substitute for m, x1, and y1.
y + 4 = – 2x + 6
Distributive property
y = – 2x + 2
= –2
Write in slope-intercept form.
1)
(-1, -6) and (2, 6)
2)
(0, 5) and (3, 1)
3)
(3, 5) and (6, 6)
4)
(0, -7) and (4, 25)
5)
(-1, 1) and (3, -3)
GUIDED PRACTICE
4.
Write an equation of the line that passes through
(–1, 6) and has a slope of 4.
ANSWER
5.
for Examples 2 and 3
y = 4x + 10
Write an equation of the line that passes through
(4, –2) and is parallel to the line y = 3x – 1.
ANSWER
y = 3x – 14
Write an equation of the line that passes through
(5, –2) and (2, 10) in slope intercept form
SOLUTION
The line passes through (x1, y1) = (5,–2) and
(x2, y2) = (2, 10). Find its slope.
10 – (–2)
y2 – y1
12
=
m=
=
–3
x2 – x1
2 –5
= –4
y2 – y1 = m(x – x1)
Use point-slope form.
y – 10 = – 4(x – 2)
Substitute for m, x1, and y1.
y – 10 = – 4x + 8
Distributive property
y = – 4x + 18
Write in slope-intercept form.
1)
a.
c.
Which of the following equations passes
through the points (2, 1) and (5, -2)?
y = 3/7x + 5
y = -x + 2
b. y = -x + 3
d. y = -1/3x + 3
a)
b)
c)
d)
y
y
y
y
=
=
=
=
-3x
-3x
-3x
-3x
–3
+ 17
+ 11
+5
9–5
y2 – y1
4
=
m=
=
2
x2 – x1
1 – -1
y – 9 = 2(x – 1)
y – 9 = 2x - 2
y = 2x + 7
-2x
-2x
-2x + y = 7
2x - y = -7
=2
y + 5 = ½(x – 4)
y + 5 = ½x - 2
y = ½x - 7
2y = x - 14
-x
-x
-x + 2y = -14
x - 2y = 14
Multiply everything by 2 to get rid
of the fraction
m=⅔
y – 4 = ⅔(x – 6)
y – 4 = ⅔x - 4
y = ⅔x
3y = 2x
-2x
Multiply everything by 3 to get rid
of the fraction
-2x
-2x + 3y = 0
2x - 3y = 0
EXAMPLE
2
Write an equation
in standard form of the line that
passes through (5, 4) and has a slope of –3.
SOLUTION
y – y1 = m(x – x1)
Use point-slope form.
y – 4 = –3(x – 5)
Substitute for m, x1, and y1.
y – 4 = –3x + 15
Distributive property
y = –3x + 19
+3x
+3x
3x + y = 19
Write in slope-intercept form.
Parallel vs. Perpendicular Lines
EXAMPLE 3
Write equations of parallel or perpendicular l
b. A line perpendicular to a line with slope m1 = –4
has a slope of m21= –1
= . Use point-slope form
m1 4
with
(x1, y1) = (–2, 3)
y – y1 = m2(x – x1)
1
y–3=
4
1
y–3=
4
1
y–3=
4
Use point-slope form.
(x – (–2))
Substitute for m2, x1, and y1.
(x +2)
Simplify.
1
2
x+
1
7
y  x
4
2
Distributive property
Write in slope-intercept form.
y = 3 (or any number)
Lines that are horizontal have a slope of zero.
They have “run” but no “rise”. The rise/run
formula for slope always equals zero since rise
= o.
y = mx + b
y = 0x + 3
y=3
This equation also describes what is happening
to the y-coordinates on the line. In this case,
they are always 3.
x = -2
Lines that are vertical have no slope
(it does not exist).
They have “rise”, but no “run”. The rise/run
formula for slope always has a zero
denominator and is undefined.
These lines are described by what is happening
to their x-coordinates. In this example, the xcoordinates are always equal to -2.
a)
b)
c)
d)
x = -5
y=7
x=y
x+y=0
a)
b)
c)
d)
Y = 2x + 3
Y – 2x = 4
2x – y = 8
Y = -2x + 1

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