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EXAMPLE 4

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

jolene-rasmussen
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EXAMPLE 4

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  1. 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 PQOwith 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). EXAMPLE 4 Apply variable coordinates

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

  3. Write a coordinate proof of the Midsegment Theorem for one midsegment. E(q+p, r) D(q, r) = = GIVEN : PROVE : SOLUTION 1 DE is a midsegment of OBC. 2 DE OCand DE = OC STEP1 Place OBCand assign coordinates. Because you are finding midpoints, use 2p, 2q, and 2r. Then find the coordinates of Dand E. 2q + 2p, 2r + 0 2q + 0, 2r + 0 E( ) D( ) 2 2 2 2 EXAMPLE 5 Prove the Midsegment Theorem

  4. STEP 2 Prove DE OC. The y-coordinates of Dand Eare the same, so DEhas a slope of 0. OCis on the x-axis, so its slope is 0. 1 Because their slopes are the same, DE OC. 2 Prove DE = OC. Use the Ruler Postulate to find DEand OC. STEP3 = 2p 2p – 0 (q +p) – q OC= = p DE= So, the length of DEis half the length of OC EXAMPLE 5 Prove the Midsegment Theorem

  5. In Example 5, find the coordinates of F, the midpoint of OC. Then show that EF OB. SOLUTION Given:FE is a midsegment. Prove:FE OB The midpoints are E (q + p, r ) and F = F (p, 0). The slope of both FEand OBis so EF ||OB r q 1 Also, FE = q2 + r2and OB = 2 q2 + r2, so FE = OB. 2 for Examples 4 and 5 GUIDED PRACTICE 7.

  6. Graph the points O(0, 0), H(m, n), and J(m, 0). Is OHJa right triangle? Find the side lengths and the coordinates of the midpoint of each side. 8. J(m,0) H(m,n) STEP1 Place OHJwith the right angle at the origin The vertices are O(0,0) , , ( 0,0) (m,n) (m,0) O H J for Examples 4 and 5 GUIDED PRACTICE SOLUTION

  7. = = = HJ = OH = = n m 2 0 + m , n +0 n C( ) C( , ) = 2 2 2 2 2 2 2 2 2 2 2 2 2 0 + (– n) m + (– n) Yes, OHJa right triangle m + n (0 – m) + (0 –n) (m – m) + (0 –n) for Examples 4 and 5 GUIDED PRACTICE STEP2 Use the distance formula to find OH and HJ Use the Midpoint Formula to find the midpoint Cof the hypotenuse.

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