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












0












$begingroup$


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?






geometry







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  • $begingroup$
    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
















0












$begingroup$


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?






geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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?






geometry







share|cite|improve this question









$endgroup$




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?






geometry




geometry






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share|cite|improve this question










asked Jan 10 at 10:23









Vali ROVali RO

736




736












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $iin {A,B,C}$.



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.






share|cite|improve this answer









$endgroup$













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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $iin {A,B,C}$.



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.






share|cite|improve this answer









$endgroup$













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
















1












$begingroup$

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 $iin {A,B,C}$.



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.






share|cite|improve this answer









$endgroup$













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














1












1








1





$begingroup$

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 $iin {A,B,C}$.



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.






share|cite|improve this answer









$endgroup$



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 $iin {A,B,C}$.



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 12:25









Siong Thye GohSiong Thye Goh

101k1466118




101k1466118












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


















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
















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




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


















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