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Three points $(1, 2), (3, 4), (5, 6)$ on the school graph paper are:
AVertices of a right triangle on the chart at the origin
BVertices of an equilateral triangle on the chart with side $2$
CNon-collinear, since they have distinct $x$-coordinates on chart
DCollinear, since the area determinant evaluates to zero
Answer & Solution
Correct answer: D. Collinear, since the area determinant evaluates to zero
$\tfrac 1 2|1(4-6) + 3(6-2) + 5(2-4)| = \tfrac 1 2|-2 + 12 - 10| = 0$; collinear.
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