incenter point coordinates given the coordinates of the three vertices of a triangle ABC

I have the following equations: $4x-y+2=0$ , $x-4y-8=0$ , $x+4y-8=0$.These equations determine a triangle.I have to find the incenter coordinates.

I found the coordinates of the triangle vertices and all I know is that I take the incenter point $I(a,b)$ then $\frac{\left | 4a-b+2 \right |}{\sqrt{17}} = \frac{\left | a-4b-8 \right |}{\sqrt{17}} = \frac{\left | a+4b-8 \right |}{\sqrt{17}}$

How to continue?

• Hint: The internal bisector of the angle at (8, 0) is the x-axis. So b = 0. – Michael Behrend Jan 10 at 12:20

Guide:

It is know that for a $$\triangle ABC$$, suppose its length is $$a,b,c$$, with the vertices being $$(x_i, y_i)$$ where $$i\in \{A,B,C\}$$.

Then the formula is given by

$$\left( \frac{ax_A+bx_B+cx_C}{a+b+c},\frac{ay_A+by_B+cy_C}{a+b+c}\right)$$

A proof of the formula can be found here.

You have found the coordinates, hence it should be possible for you to find the lenght of the sides easily.

• Thanks a lot!I did it. – Vali RO Jan 10 at 14:25