A question on in-centres and ex-centres.

Question

$I$, $I_1$, $I_2$, $I_3$ are the in-centres and ex-centres of $\triangle ABC$. If $I(0,0)$, $I_1(2,3)$, $I_2(5,7)$ then the distance between the orthocentres of $\triangle II_1I_3$ and $\triangle I_1I_2I_3$ is?

I tried using the in-centres and ex-centres formulae but again ended up with complications. Can anyone please help me out with a simple solution? Also, since the origin is in-centre, can we tell it's an equilateral triangle?

P.S.- it's my first question using MathJax so don't downvote my question for any mistakes

Answer 1

You have to use an important property of the incenter and the three excenters of a triangle, that these four points form a so-called orthocentric system (see, for example, this link and then this one). I quote: "If four points form an orthocentric system, then each of the four points is the orthocenter of the other three."
Therefore, the orthocenter of $\bigtriangleup II_1I_3$ is $I_2$ and the orthocenter of $\bigtriangleup I_1I_2I_3$ is $I$. The requested distance between the two orthocenters is then $I_2I=\sqrt{(5-0)^2+(7-0)^2}=\sqrt{74}$.

Answer 2

It's just $II_2=\sqrt{5^2+7^2}=\sqrt{74}$.