1. Lesson 5.1, For use with pages 295-301
In Exercises 1– 4, use A(0, 10), B(24, 0), and C(0, 0).
1. Find AB.
ANSWER 26
2. Find the midpoint of CA.
ANSWER
(0, 5)

2. Lesson 5.1, For use with pages 295-301
In Exercises 1– 4, use A(0, 10), B(24, 0), and C(0, 0).
3. Find the midpoint of AB.
ANSWER
(12, 5)
4. Find the slope of AB.
ANSWER – 5 12

4. EXAMPLE 1
Use the Midsegment Theorem to find lengths
Triangles are used for strength in roof trusses. In the diagram, UV and VW are midsegments of
Find UV and RS.
RST.
CONSTRUCTION
SOLUTION UV = 1 2 RT
( 90 in.)
45 in. RS = 2 VW
( 57 in.)
114 in.

5. GUIDED PRACTICE
for Example 1 1.
Copy the diagram in Example 1. Draw and name the third midsegment.
ANSWER UW 2.
In Example 1, suppose the distance UW is 81 inches. Find VS.
ANSWER
81 in.

6. EXAMPLE 2
Use the Midsegment Theorem
In the kaleidoscope image, AE BE and AD CD . Show that CB DE .
SOLUTION
Because AE BE and AD CD , E is the midpoint of AB and D is the midpoint of AC by definition.
Then DE is a midsegment of ABC by definition and CB DE by the Midsegment Theorem.

8. EXAMPLE 3
Place a figure in a coordinate plane
Place each figure in a coordinate plane in a way that is convenient for finding side lengths. Assign coordinates to each vertex. a.
A rectangle b.
A scalene triangle
SOLUTION
It is easy to find lengths of horizontal and vertical segments and distances from (0, 0), so place one vertex at the origin and one or more sides on an axis.

9. EXAMPLE 3
Place a figure in a coordinate plane a.
Let h represent the length and k represent the width. b.
Notice that you need to use three different variables.

10. Homework
298: 1-11

11. GUIDED PRACTICE
for Examples 2 and 3 3.
In Example 2, if F is the midpoint of CB , what do you know about DF ?
ANSWER
DF is a midsegment of ABC.
DF AB and DF is half the length of AB. 4.
Show another way to place the rectangle in part (a) of Example 3 that is convenient for finding side lengths. Assign new coordinates.
ANSWER

12. GUIDED PRACTICE
for Examples 2 and 3 5.
Is it possible to find any of the side lengths in part (b) of Example 3 without using the Distance Formula? Explain.
Yes; the length of one side is d.
ANSWER 6.
A square has vertices (0, 0), (m, 0), and (0, m). Find the fourth vertex.
ANSWER
(m, m)

13. EXAMPLE 4
Apply variable coordinates
Place an isosceles right triangle in a coordinate plane. Then find the length of the hypotenuse and the coordinates of its midpoint M.
SOLUTION
Place PQO with the right angle at the origin. Let the length of the legs be k. Then the vertices are located at P(0, k), Q(k, 0), and O(0, 0).

14. Apply variable coordinates
EXAMPLE 4
Apply variable coordinates
Use the Distance Formula to find PQ.
PQ =
(k – 0) + (0 – k) 2 =
k + (– k) 2 =
k + k 2 = 2k 2
= k 2
Use the Midpoint Formula to find the midpoint M of the hypotenuse.
M( )
0 + k , k + 0 2 =
M( , ) k 2

15. EXAMPLE 5
Prove the Midsegment Theorem
Write a coordinate proof of the Midsegment Theorem for one midsegment.
GIVEN :
DE is a midsegment of OBC.
PROVE :
DE OC and DE = OC 1 2
SOLUTION
STEP 1
Place OBC and assign coordinates. Because you are finding midpoints, use 2p, 2q, and 2r. Then find the coordinates of D and E.
D( )
2q + 0, 2r + 0 2 =
D(q, r)
E( )
2q + 2p, 2r + 0
E(q+p, r)

16. EXAMPLE 5
Prove the Midsegment Theorem
STEP 2
Prove DE OC . The y-coordinates of D and E are the same, so DE has a slope of 0. OC is on the x-axis, so its slope is 0.
Because their slopes are the same, DE OC .
STEP 3
Prove DE = OC. Use the Ruler Postulate 1 2
to find DE and OC .
DE =
(q + p) – q
= p
OC =
2p – 0
= 2p
So, the length of DE is half the length of OC

17. GUIDED PRACTICE
for Examples 4 and 5 7.
In Example 5, find the coordinates of F, the midpoint of OC . Then show that EF OB .
(p, 0); slope of EF = = ,
slope of OB = = , the slopes of
EF and OB are both , making EF || OB.
r  0
(q + p)  p q r
2r  0
2q  0
ANSWER

18. GUIDED PRACTICE
for Examples 4 and 5 8.
Graph the points O(0, 0), H(m, n), and J(m, 0). Is OHJ a right triangle? Find the side lengths and the coordinates of the midpoint of each side.
ANSWER
yes; OJ = m, JH = n,
HO = m2 + n2,
OJ: ( , 0), JH: (m, ),
HO: ( , ) 2 m n
Sample: