Let $A$ the triangle formed by the vertices $(x₁,y₁),(x₂,y₂),(x₃,y₃).$

Find sufficient and necessary conditions for which the area of $A$ is zero.

If the vertices $(x₁,y₁),(x₂,y₂),(x₃,y₃)$ are equal, then the triangle will shrink to a single point and hence its area is zero, but this is not the general case.

  • $\begingroup$ That's a very simple case in which the area is zero. Can't you think of a less trivial case? Then the rest should be fairly obvious, or something you can easily get help on. $\endgroup$ – Henrik Feb 16 '15 at 8:26
  • $\begingroup$ @Henrik: How we can explain this geometrically. $\endgroup$ – DER Feb 16 '15 at 8:30

Fix one of the vertices and form the vectors describing two sides: for example $u (x_2-x_1, y_2-y_1)$ and $v = (x_3-x_1, y_3-y_1)$. The triangle will have area zero if and only if $u$ and $v$ are linearly dependent, i.e. if and only if $$ \det \begin{bmatrix} x_2-x_1 & y_2-y_1 \\ x_3-x_1 & y_3-y_1 \end{bmatrix} = 0. $$

  • $\begingroup$ @ mrf: How we can explain this geometrically $\endgroup$ – DER Feb 16 '15 at 8:32
  • 1
    $\begingroup$ All three points will be collinear. $\endgroup$ – mrf Feb 16 '15 at 8:35

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