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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Homer on April 17, 2013, 04:59:47 pm

Title: :(
Post by: Homer on April 17, 2013, 04:59:47 pm: :(
Title: Re: :(
Post by: sin0001 on April 17, 2013, 05:04:12 pm: Let AB---> a and BC----> b
AC = AB + BC = a + b
DE = DB + BE = 1/2AB + 1/2BC = 1/2 (a + b)
Therefore DE = 1/2 AC, hence DE//AC
Title: Re: :(
Post by: TheManG on April 17, 2013, 05:10:56 pm: $\overrightarrow{AB}=\underline{a}$
$\overrightarrow{CB}=\underline{b}$

$\overrightarrow{AD}=\frac{1}{2}\overrightarrow{AB}$
$\overrightarrow{AD}=\frac{1}{2}\underline{a}$

$\overrightarrow{CE}=\frac{1}{2}\overrightarrow{CB}$
$\overrightarrow{AD}=\frac{1}{2}\underline{b}$

$\overrightarrow{AC}=\overrightarrow{AB}-\overrightarrow{CB}$
$\overrightarrow{AC}=\underline{a}-\underline{b}$

$\overrightarrow{DE}=\overrightarrow{DB}-\overrightarrow{BE}$
$\overrightarrow{AC}=\frac{1}{2}\underline{a}-\frac{1}{2}\underline{b}$

( $\overrightarrow{DB}=\frac{1}{2}\overrightarrow{AB} and \overrightarrow{BE}=-\frac{1}{2}\overrightarrow{CB}$

$\overrightarrow{DB}=\frac{1}{2}(\overrightarrow{DB}-\overrightarrow{BE}$
$\overrightarrow{DB}=\frac{1}{2}(\underline{a}-\underline{b})$
$\overrightarrow{DB}=\frac{1}{2}(\overrightarrow{AC})$

(Could have errors as I did this in the 'moment'. i.e No paper with me just everything typed up)
Title: Re: :(
Post by: Homer on April 17, 2013, 06:56:56 pm: thanks guys :)