1. 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).

2. 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

3. 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)

4. 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

5. 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

6. 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: