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A B points in plane. M symetric of A face of B , N symetric of B face of A. show MN...
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The problem provides the information that M is symmetric of A with respect to B, hence, B is the midpoint of line segment AM.
Using the position vectors `bar r_A, bar r_B, bar r_M` , yields:
`bar r_B = (1/2)(bar r_A + bar r_M) ` `=> bar r_M = 2bar r_B - bar r_A`
The problem provides the information that N is symmetric of B with respect to A, hence, A is the midpoint of line segment BN.
`bar r_A = (1/2)(bar r_B + bar r_N) => bar r_N = 2bar r_A - bar r_B`
Supposing that P is the midpoint of AB yields:
`bar r_P = (1/2)(bar r_A + bar r_B)`
Supposing that P' is the midpoint of MN yields:
`bar r_P' = (1/2)(bar r_M + bar r_N)`
Replacing `2bar r_B - bar r_A` for `bar r_M` and `2bar r_A - bar r_B` for `bar r_N` , yields:
`bar r_P' = (1/2)(2bar r_B - bar r_A + 2bar r_A - bar r_B)`
`bar r_P' = (1/2)(bar r_A + bar r_B) = bar r_P`
Hence, the midpoint P coincides with the midpoint P', hence, the line segments MN and AB share the midpoint P.
Posted by sciencesolve on May 3, 2013 at 6:16 PM (Answer #1)
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