EXAMPLE 1 Find segment lengths Skateboard In the skateboard design, VW bisects XY at point T, and XT = 39.9 cm.
Download ReportTranscript EXAMPLE 1 Find segment lengths Skateboard In the skateboard design, VW bisects XY at point T, and XT = 39.9 cm.
Slide 1
EXAMPLE 1
Find segment lengths
Skateboard
In the skateboard design, VW bisects XY at point T,
and XT = 39.9 cm. Find XY.
SOLUTION
Point T is the midpoint of XY .
So, XT = TY = 39.9 cm.
XY = XT + TY
Segment Addition Postulate
= 39.9 + 39.9 Substitute.
Add.
= 79.8 cm
Slide 2
EXAMPLE 2
Use algebra with segment lengths
ALGEBRA Point M is the midpoint
of VW . Find the length of VM .
SOLUTION
STEP 1
Write and solve an equation. Use the fact
that that VM = MW.
VM = MW
4x – 1 = 3x + 3
x–1=3
x=4
Write equation.
Substitute.
Subtract 3x from each side.
Add 1 to each side.
Slide 3
EXAMPLE 2
STEP 2
Use algebra with segment lengths
Evaluate the expression
for VM when x = 4.
VM = 4x – 1 = 4(4) – 1 = 15
So, the length of VM is 15.
Check: Because VM = MW, the length of MW should be
15. If you evaluate the expression for MW, you should
find that MW = 15.
MW = 3x + 3 = 3(4) +3 = 15
Slide 4
GUIDED PRACTICE
1.
for Examples 1 and 2
In Exercises 1 and 2, identify the segment bisector
of PQ . Then find PQ.
SOLUTION
M is midpoint and line MN bisects the line PQ at M.
7
So MN is the segment bisector of PQ. So PM = MQ =1 8
PQ = PM + MQ
= 1 78 + 1 78
3
=3 4
Segment addition postulate.
Substitute
Add.
Slide 5
GUIDED PRACTICE
2.
for Examples 1 and 2
In Exercises 1 and 2, identify the segment bisector of
PQ . Then find PQ.
SOLUTION
M is midpoint and line l bisects the line PQ of M. So l
is the segment bisector of PQ. So PM = MQ
Slide 6
GUIDED PRACTICE
for Examples 1 and 2
STEP 1 Write and solve an equation
PM = MQ
5x – 7 = 11 – 2x
7x = 18
18
x=
7
Write equation.
Substitute.
Add 2x and 7 each side.
Divide each side by 7.
18
STEP 2 Evaluate the expression for PQ when x = 7
PQ = 5x – 7 + 11 – 2x = 3x + 4
Substitute 18 for x.
PQ = 3 18 + 4
7
7
5
Simplify.
11
=
7
Slide 7
EXAMPLE
3
a.
Use the Midpoint Formula
FIND MIDPOINT The endpoints of RS are
R(1,–3) and S(4, 2). Find the coordinates of
the midpoint M.
Slide 8
EXAMPLE Use the Midpoint Formula
3
SOLUTION
a.
FIND MIDPOINT Use the Midpoint
Formula.
,– 3 + 2 = M 5,– 1
M1+
2
2 2
42
ANSWER The coordinates of the midpoint M
are 5 ,– 1
2 2
Slide 9
EXAMPLE Use the Midpoint Formula
3 b. FIND ENDPOINT The midpoint of JK is M(2,
1). One endpoint is J(1, 4). Find the
coordinates of endpoint K.
FIND ENDPOINT Let (x, y) be the
coordinates of endpoint K. Use
the Midpoint Formula.
STEP 1 Find x.
1+ x = 2
2
1+x=4
x=3
STEP 2 Find y.
4+ y 1
2 =
4+y=2
y=–
2
ANSWER The coordinates of endpoint K are (3, – 2).
Slide 10
GUIDED PRACTICE
for Example 3
The endpoints of AB are A(1, 2) and B(7,
8).Find the coordinates of the midpoint M.
SOLUTION
3.
Use the midpoint
formula.
1 + ,2 + 8
M
(4, 5)
=
M
2
72
(
)
ANSWER The Coordinates of the midpoint M
are (4,5).
Slide 11
GUIDED PRACTICE for Example 3
4.
The midpoint of VW is M(– 1, – 2). One
endpoint is W(4, 4). Find the coordinates
of endpoint V.
SOLUTION
Let (x, y) be the coordinates of endpoint V.
Use the Midpoint Formula.
STEP 1 Find x.
STEP 2 Find y.
4+ x = – 1
2
4+x=–2
x=–6
4+ y – 2
2 =
4+y=–4
y=–8
The coordinates of endpoint V is (– 6, – 8)
Slide 12
EXAMPLE 4 Standardized Test Practice
SOLUTION
Use the Distance Formula. You
may find it helpful to draw a
diagram.
Slide 13
EXAMPLE 4 Standardized Test Practice
2
2
RS = (x2 – x1 ) + (y 2 – y1 )
=
=
=
=
2
[(4 – 2)] + [(–1) –3]
2
2
(2) + (–4 )
4+16
20
= 4.47
Distance Formula
2
Substitute.
Subtract.
Evaluate powers.
Add.
Use a calculator to approximate
the square root.
The correct answer is C.
Slide 14
GUIDED PRACTICE for Example 4
5.
In Example 4, does it matter which ordered
pair you choose to substitute for (x1 , y1) and
which ordered pair you choose to substitute
for (x2 , y2 )? Explain.
ANSWER
No, when squaring the difference in the
coordinate you get the same answer as long as
you choose the x and y value from the some
period
Slide 15
GUIDED PRACTICE for Example 4
6.
What is the approximate length of AB , with
endpoints A(–3, 2) and B(1, –4)?
6.1 units
7.2 units
8.5 units
10.0 units
SOLUTION
Use the Distance Formula. You may find it helpful
to draw a diagram.
A =
B
=
=
2
2
(x2 – x1 ) + (y 2 – y1 )
2
[2 –(–3)] + (–4 –1)
2
(5) + (5 )
2
Distance Formula
2
Substitute.
Subtract.
Slide 16
GUIDED PRACTICE for Example 4
= 25+25
Evaluate powers.
=
Add.
50
= 7.2
ANSWER
Use a calculator to approximate
the square root.
The correct answer is B
EXAMPLE 1
Find segment lengths
Skateboard
In the skateboard design, VW bisects XY at point T,
and XT = 39.9 cm. Find XY.
SOLUTION
Point T is the midpoint of XY .
So, XT = TY = 39.9 cm.
XY = XT + TY
Segment Addition Postulate
= 39.9 + 39.9 Substitute.
Add.
= 79.8 cm
Slide 2
EXAMPLE 2
Use algebra with segment lengths
ALGEBRA Point M is the midpoint
of VW . Find the length of VM .
SOLUTION
STEP 1
Write and solve an equation. Use the fact
that that VM = MW.
VM = MW
4x – 1 = 3x + 3
x–1=3
x=4
Write equation.
Substitute.
Subtract 3x from each side.
Add 1 to each side.
Slide 3
EXAMPLE 2
STEP 2
Use algebra with segment lengths
Evaluate the expression
for VM when x = 4.
VM = 4x – 1 = 4(4) – 1 = 15
So, the length of VM is 15.
Check: Because VM = MW, the length of MW should be
15. If you evaluate the expression for MW, you should
find that MW = 15.
MW = 3x + 3 = 3(4) +3 = 15
Slide 4
GUIDED PRACTICE
1.
for Examples 1 and 2
In Exercises 1 and 2, identify the segment bisector
of PQ . Then find PQ.
SOLUTION
M is midpoint and line MN bisects the line PQ at M.
7
So MN is the segment bisector of PQ. So PM = MQ =1 8
PQ = PM + MQ
= 1 78 + 1 78
3
=3 4
Segment addition postulate.
Substitute
Add.
Slide 5
GUIDED PRACTICE
2.
for Examples 1 and 2
In Exercises 1 and 2, identify the segment bisector of
PQ . Then find PQ.
SOLUTION
M is midpoint and line l bisects the line PQ of M. So l
is the segment bisector of PQ. So PM = MQ
Slide 6
GUIDED PRACTICE
for Examples 1 and 2
STEP 1 Write and solve an equation
PM = MQ
5x – 7 = 11 – 2x
7x = 18
18
x=
7
Write equation.
Substitute.
Add 2x and 7 each side.
Divide each side by 7.
18
STEP 2 Evaluate the expression for PQ when x = 7
PQ = 5x – 7 + 11 – 2x = 3x + 4
Substitute 18 for x.
PQ = 3 18 + 4
7
7
5
Simplify.
11
=
7
Slide 7
EXAMPLE
3
a.
Use the Midpoint Formula
FIND MIDPOINT The endpoints of RS are
R(1,–3) and S(4, 2). Find the coordinates of
the midpoint M.
Slide 8
EXAMPLE Use the Midpoint Formula
3
SOLUTION
a.
FIND MIDPOINT Use the Midpoint
Formula.
,– 3 + 2 = M 5,– 1
M1+
2
2 2
42
ANSWER The coordinates of the midpoint M
are 5 ,– 1
2 2
Slide 9
EXAMPLE Use the Midpoint Formula
3 b. FIND ENDPOINT The midpoint of JK is M(2,
1). One endpoint is J(1, 4). Find the
coordinates of endpoint K.
FIND ENDPOINT Let (x, y) be the
coordinates of endpoint K. Use
the Midpoint Formula.
STEP 1 Find x.
1+ x = 2
2
1+x=4
x=3
STEP 2 Find y.
4+ y 1
2 =
4+y=2
y=–
2
ANSWER The coordinates of endpoint K are (3, – 2).
Slide 10
GUIDED PRACTICE
for Example 3
The endpoints of AB are A(1, 2) and B(7,
8).Find the coordinates of the midpoint M.
SOLUTION
3.
Use the midpoint
formula.
1 + ,2 + 8
M
(4, 5)
=
M
2
72
(
)
ANSWER The Coordinates of the midpoint M
are (4,5).
Slide 11
GUIDED PRACTICE for Example 3
4.
The midpoint of VW is M(– 1, – 2). One
endpoint is W(4, 4). Find the coordinates
of endpoint V.
SOLUTION
Let (x, y) be the coordinates of endpoint V.
Use the Midpoint Formula.
STEP 1 Find x.
STEP 2 Find y.
4+ x = – 1
2
4+x=–2
x=–6
4+ y – 2
2 =
4+y=–4
y=–8
The coordinates of endpoint V is (– 6, – 8)
Slide 12
EXAMPLE 4 Standardized Test Practice
SOLUTION
Use the Distance Formula. You
may find it helpful to draw a
diagram.
Slide 13
EXAMPLE 4 Standardized Test Practice
2
2
RS = (x2 – x1 ) + (y 2 – y1 )
=
=
=
=
2
[(4 – 2)] + [(–1) –3]
2
2
(2) + (–4 )
4+16
20
= 4.47
Distance Formula
2
Substitute.
Subtract.
Evaluate powers.
Add.
Use a calculator to approximate
the square root.
The correct answer is C.
Slide 14
GUIDED PRACTICE for Example 4
5.
In Example 4, does it matter which ordered
pair you choose to substitute for (x1 , y1) and
which ordered pair you choose to substitute
for (x2 , y2 )? Explain.
ANSWER
No, when squaring the difference in the
coordinate you get the same answer as long as
you choose the x and y value from the some
period
Slide 15
GUIDED PRACTICE for Example 4
6.
What is the approximate length of AB , with
endpoints A(–3, 2) and B(1, –4)?
6.1 units
7.2 units
8.5 units
10.0 units
SOLUTION
Use the Distance Formula. You may find it helpful
to draw a diagram.
A =
B
=
=
2
2
(x2 – x1 ) + (y 2 – y1 )
2
[2 –(–3)] + (–4 –1)
2
(5) + (5 )
2
Distance Formula
2
Substitute.
Subtract.
Slide 16
GUIDED PRACTICE for Example 4
= 25+25
Evaluate powers.
=
Add.
50
= 7.2
ANSWER
Use a calculator to approximate
the square root.
The correct answer is B