Download Additional Examples Lesson 3-6 Additional Examples Lesson 3-5

Additional Examples
5
EXAMPLE
A regular hexagon is inscribed in a rectangle.
Explain how you know that all the angles labeled l1 have
equal measures.
1
2
2
2
2
1
1
© Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved.
Lesson 3-5
1
Sample: The hexagon is regular, so all its angles are congruent. An
exterior angle is the supplement of a polygon’s angle because they
are adjacent angles that form a straight angle. Because supplements
of congruent angles are congruent, all the angles marked l1 have
equal measures.
Additional Examples
1
Lesson 3-6
EXAMPLE
Use the slope and y-intercept to graph the line
y ≠ –2x ± 9.
When an equation is written in the form y ≠ mx ± b, m is the slope
and b is the y-intercept. In the equation y ≠ –2x ± 9, the slope is –2
and the y-intercept is 9.
9
Move down 8
2 units.
7
6
5
4
3
2
1
5
y
Start at (0, 9).
Move right
1 unit.
x
3 1 0 1 2 3 4 5 6
1
Geometry, Chapter 3
46
Additional Examples
2
EXAMPLE
Use the x-intercept and y-intercept to graph
5x – 6y ≠ 30.
To find the x-intercept, substitute
0 for y and solve for x.
© Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved.
Lesson 3-6
To find the y-intercept, substitute
0 for x and solve for y.
5x – 6y ≠ 30
5x – 6y ≠ 30
5x – 6(0) ≠ 30
5(0) – 6y ≠ 30
5x – 0 ≠ 30
0 – 6y ≠ 30
5x ≠ 30
–6y ≠ 30
x≠6
y ≠ –5
The x-intercept is 6.
A point on the line is (6, 0).
The y-intercept is –5.
A point on the line is (0, –5).
Plot (6, 0) and (0, –5). Draw the line containing the two points.
y
10
8
6
4
2
(6, 0)
x
10 6 2 0 2 4 6 8 10
4
8
10
(0, 5)
Geometry, Chapter 3
47
Additional Examples
3
Lesson 3-6
Transform the equation –6x ± 3y ≠ 12 to slopeintercept form, then graph the resulting equation.
EXAMPLE
Step 1: Transform the equation to slope-intercept form.
–6x ± 3y ≠ 12
3y ≠ 6x ± 12
3y
6x
12
3 5 3 1 3
Add 6x to each side.
Divide each side by 3.
y ≠ 2x ± 4
© Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved.
The y-intercept is 4 and the slope is 2.
Step 2: Use the y-intercept and the slope to plot two points
and draw the line containing them.
6
y
(1, 6)
4 (0, 4)
2
–2
0
2
4x
–2
Geometry, Chapter 3
48
Additional Examples
4
Lesson 3-6
EXAMPLE
Write an equation in point-slope form of the line
with slope –8 that contains P(3, –6).
y – y1 ≠ m(x – x1) Use point-slope form.
y – (–6) ≠ –8(x – 3) Substitute –8 for m and (3, –6) for (x1, y1).
y ± 6 ≠ –8(x – 3)
© Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved.
5
Simplify.
EXAMPLE
Write an equation in point-slope form of the line
that contains the points G(4, –9) and H(–1, 1).
Step 1: Find the slope.
y 2y
m 5 x22 2 x11
1 2 ( 29)
Use the formula for slope.
m 5 21 2 4
Substitute (4, –9) for (x1, y1) and (–1, 1) for (x2, y2).
10
m52
5
m ≠ –2
Simplify.
Step 2: Select one of the points. Write the equation in
point-slope form.
y – y1 ≠ m(x – x1)
y – (–9) ≠ –2(x – 4)
y ± 9 ≠ –2(x – 4)
6
Point-slope form
Substitute –2 for m and (4, –9) for (x1, y1).
Simplify.
EXAMPLE
Write equations for the horizontal line and the
vertical line that contain A(–7, –5).
Every point on the horizontal line through A(–7, –5) has the same
y-coordinate, –5, as point A. The equation of the line is y ≠ –5. It
crosses the y-axis at (0, –5).
Every point on the vertical line through A(–7, –5) has the same
x-coordinate, –7, as point A. The equation of the line is x ≠ –7. It
crosses the x-axis at (–7, 0).
Geometry, Chapter 3
49