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.

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Presentation transcript:

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) 2 2 = k + (– k) 22 = k + k 22 = 2k2k 2 = k 2 Use the Midpoint Formula to find the midpoint M of the hypotenuse. M( ) 0 + k, k = M(, ) k 2 k 2

EXAMPLE 5 Prove the Midsegment Theorem GIVEN : DE is a midsegment of OBC. PROVE : DE OC and DE = OC 1 2 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 = D(q, r) E( ) 2q + 2p, 2r = 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 Postulate 1 2 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 so q r EF ||OB Also, FE = q 2 + r 2 and OB = 2 q 2 + r 2, so FE = OB. 1 2

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) 2 2 = m + (– n) 22 = m + n 22 Use the Midpoint Formula to find the midpoint C of the hypotenuse. C( ) 0 + m, n = C(, ) m 2 n 2 HJ = (m – m) + (0 – n) 2 2 = 0 + (– n) 22 = n Yes, OHJ a right triangle