# Show algebraically that a triangle cannot be drawn with vertices at the three points A(2, 5), B(6, 1), and C(8, -1). There are (at least) two distinct approaches to this problem.

Show algebraically that a triangle cannot be drawn with vertices at the three points A(2, 5), B(6, 1), and C(8, -1). There are (at least) two distinct approaches to this problem.
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grocbyntza
A(2,5), B(6,1), C(8,-1)
like equation of AB
$⇒y-5=\frac{4}{-4}\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒y-5=x+2\phantom{\rule{0ex}{0ex}}⇒x+y-7=0$
for point C (8,-1)
$⇒8-1-7=0$
All three points are on line line $\mathrm{△}$ formation is not posible.
$\stackrel{\to }{AB}=4i-4i=4\phantom{\rule{0ex}{0ex}}\stackrel{\to }{AC}=6i-6i$
Both $\stackrel{\to }{AB}$ and $\stackrel{\to }{AC}$ are parallel to each other