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Given the triangle $ABC$ with its vertices $A(0,1)$, $B(-2,1)$, $C(8,-8)$. Determine the intersection point of the median $AM$ and the line $l$, if $l\parallel AB$ and $C$ is element of $l$.

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    "verticles"? : ) I guess you will have to apply someone's seminal work to solve this problem.2012-12-13

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$M(\frac{-2+8}2,\frac{1-8}2)$ i.e., $M(3,-\frac72)$

So, $AM: \frac{y-1}{x-0}=\frac{1-\frac{-7}2}{0-3}\implies 3x+2y-2=0--->(1)$

$AB: \frac{y-1}{x-0}=\frac{1-1}{0-(-2)}\implies y=1$

So, any line parallel to $AB$ will be $y=a$ where $a$ is some constant.

As $l,$ passes through $C(8,-8),a=-8\implies l:y=-8--->(2)$

Now, find the intersection of $AM, l$

Putting $y=-8$ in $(1), x=\frac{2-2(-8)}3=6$

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Let $E$ be the intersection. Use angle-side-angle you will find that $M$ is the midpoint of the $E$ and $A$. In other word, we have $ABEC$ being a parallelogram. Now remember the paralellogram law of vector summation? This means $E=A+(C-A)+(B-A)=B+C-A$.