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I came up with this question when I was trying to figure out the coordinates of the incenter of a triangle with equations: $4x-3y=0$, $3x-4y+12=0$, $3x+4y+2=0$.

I assumed the coordinates of the incenter to be $(h,k)$ and equated the perpendicular distances from all the sides and got:

$4h-3k=\pm(3h-4k+12)$, $3h-4k+12=\pm(3h+4k+2)$, $3h+4k+2=\pm(4h-3k) $

But I didn't know which sign to take. As different signs would mean different angle bisectors(external or internal). Image showing intersection of Angle bisectors(posted as a link due to low reputation restrictions)

I found this Can we find incentre of a triangle by using equation of lines?, answering which bisector to take but I couldn't understand the method in the case when I do not have the vertices.

P.S. I want to know the incenter without knowing the vertices. Also I am curious why the above method works, intuitively. Thanks.

asks281
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  • If you draw the normal vectors of the lines, as given in the equation (e.g. for $4x-3y=0$ it is the vector $(4,-3)$, then choose its sign according to whether it is pointing towards the internal of the triangle or pointing outside. – Berci Oct 27 '19 at 13:17
  • Could you elaborate please? – asks281 Oct 27 '19 at 18:47
  • I can’t think of a good way to do this purely analytically, since you don’t necessarily want the acute bisector every time. There is a way to adjust the three normals so that they point either consistently inward or outward (though you won’t be able to tell which) so that you can tell which of the four circles tangent to the three lines is the incircle, but I think that’s more work than simply computing the vertices. Each pairwise intersection only requires a cross product and a couple of divisions, after all. – amd Oct 30 '19 at 21:55

1 Answers1

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$4·x-3·y=0$

$3·x-4·y+12=0$

$3·x+4·y+2=0$

multiply the equations by the signs of the cofactors of the constants: 0, 12 and 2.

$M=\left [ \begin{array}{} 4 & -3 & 0 \\ 3 & -4 & 12 \\ 3 & 4 & 2\\ \end{array} \right ] $

$cofactor{M_{13}}=\left|\begin{array}{} 3 & -4 \\ 3 & 4 \\ \end{array} \right| =24$

$cofactor{M_{23}}=-\left|\begin{array}{} 4 & -3 \\ 3 & 4 \\ \end{array} \right| =-25$

$cofactor{M_{33}}=\left|\begin{array}{} 4 & -3 \\ 3 & -4 \\ \end{array} \right| =-7$

sign(24)=1, sign(-25)=-1 and sign(-7)=-1

$1·(4·x-3·y)=0⇒4·x-3·y=0$

$-1·(3·x-4·y+12)=0⇒-3·x+4·y-12=0$

$-1·(3·x+4·y+2)=0⇒-3·x-4·y-2=0$

with these "oriented" equations we obtain the bisectors (the three generate internal bisectors)

$\frac{4·x-3·y}{\sqrt{4^2+(-3)^2}}=\frac{-3·x-4·y-2}{\sqrt{(-3)^2+(-4)^2}}$

$\frac{-3·x+4·y-12}{\sqrt{(-3)^2+4^2}}=\frac{-3·x-4·y-2}{\sqrt{(-3)^2+(-4)^2}}$

$\frac{4·x-3·y}{\sqrt{4^2+(-3)^2}}=\frac{-3·x+4·y-12}{\sqrt{(-3)^2+4^2}}$

$7·x+y+2=0$

$4·y-5=0$

$7·x-7·y+12=0$

bisectorInternal

Wilson Massaro
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