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?

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


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.

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

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