EXAMPLE 4 Apply variable coordinates

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).

Place an isosceles right triangle in a coordinate plane. Then find the length of the hypotenuse and the coordinates of its midpoint M.

EXAMPLE 4 Apply variable coordinates

Use the Distance Formula to find PQ.

PQ = (k – 0) + (0 – k)22 = k + (– k)

2 2 = k + k2 2

= 2k 2 = k 2

Use the Midpoint Formula to find the midpoint M of the hypotenuse.

M( )0 + k , k + 02 2 = M( , )k

2k2

EXAMPLE 5 Prove the Midsegment Theorem

GIVEN : DE is a midsegment of OBC.

PROVE : DE OC and DE = OC12

Write a coordinate proof of the Midsegment Theorem for one midsegment.

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 + 02 2

= D(q, r) E( )2q + 2p, 2r + 02 2

= E(q+p, r)

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.

STEP 3 Prove DE = OC. Use the Ruler Postulate12

to find DE and OC .

DE =(q + p) – q= p OC = 2p – 0 = 2p

Because their slopes are the same, DE OC .

So, the length of DE is half the length of OC

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 .

Prove:FE OB

Given:FE is a midsegment.

SOLUTION

The midpoints are E (q + p, r ) and F = F (p, 0). The slope of both FE and OB is soq

r EF ||OB

Also, FE = q2 + r2 and OB = 2 q2 + r2, so FE = OB.12

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.

SOLUTION

J

STEP 1 Place OHJ with the right

angle at the origin

The vertices are

( 0,0) (m,0)(m,n)HO , ,

H(m,n)

O(0,0)

J(m,0)

GUIDED PRACTICE for Examples 4 and 5

STEP 2

Use the distance formula to find OH and HJ

OH = (0 – m) + (0 – n) 22 = m + (– n)

2 2 = m + n2 2

Use the Midpoint Formula to find the midpoint C of the hypotenuse.

C( )0 + m , n + 02 2 = C( , )m

2n2

HJ = (m – m) + (0 – n) 22 = 0 + (– n)

2 2= n

Yes, OHJ a right triangle