0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $$ \det \begin{bmatrix} x_2-x_1 & y_2-y_1 \\ x_3-x_1 & y_3-y_1 \end{bmatrix} = 0. $$

$\endgroup$
  • $\begingroup$ @ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.