find center of the triangle formed by the points (0,0,0),(3,0,0),and(0,4,0)
Answers
Let A(0,0), B(0,3), C(4,0) be the given points. The excenter w.r.t. A is IA(6,6) in the picture:
Excenter of the triangle with sides 3, 4, 5
First of all, it is on the angle bisector of the angle BACˆ, which is the first bisector in the coordinate system. So it is a point of the shape (s,s). Then the distance from (s,s) to the line BC given by the equation BC: 3x+4y−12=0, or better in normed form
BC : 35x+45y−125=0 ,
is obtained by plugging in (s,s) in the above "normed equation". We get (7s−12)/5. (And have to take absolute value of it.) The distance from (s,s) to the axes (which are corresponding to the other two sides) is s. So we get the equation s=(7s−12)/5. The solution is s=6.
Comment: An other way to get the touching point P of the excircle with the hypotenuse BC is to compute the half-perimeter p of the triangle ABC with sides 3,4,5, it is p=(3+4+5)/2=6. Now the point P is on BC at distances p−b=6−4=2, and p−c=6−3=3 from C, respectively B. (Together we get 2+3=5=BC.)
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