Presentation on theme: "EXAMPLE 4 Apply variable coordinates"— Presentation transcript:
1 EXAMPLE 4Apply variable coordinatesPlace an isosceles right triangle in a coordinate plane. Then find the length of the hypotenuse and the coordinates of its midpoint M.SOLUTIONPlace 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 4Apply variable coordinatesUse the Distance Formula to find PQ.PQ =(k – 0) + (0 – k)2=k + (– k)2=k + k2=2k2= k2Use the Midpoint Formula to find the midpoint M of the hypotenuse.M( )0 + k , k + 02=M( , )k2
3 EXAMPLE 5Prove the Midsegment TheoremWrite a coordinate proof of the Midsegment Theorem for one midsegment.GIVEN :DE is a midsegment of OBC.PROVE :DE OC and DE = OC12SOLUTIONSTEP 1Place 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=D(q, r)E( )2q + 2p, 2r + 0E(q+p, r)
4 EXAMPLE 5Prove the Midsegment TheoremSTEP 2Prove 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 3Prove DE = OC. Use the Ruler Postulate12to find DE and OC .DE =(q + p) – q= pOC =2p – 0= 2pSo, the length of DE is half the length of OC
5 GUIDED PRACTICEfor Examples 4 and 57.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 ofEF and OB are both , making EF || OB.r 0(q + p) pqr2r 02q 0ANSWER
6 GUIDED PRACTICEfor Examples 4 and 58.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.ANSWERyes; OJ = m, JH = n,HO = m2 + n2,OJ: ( , 0), JH: (m, ),HO: ( , )2mnSample: