The in radius of the triangle formed by the points (0,3), (4,0) and (0,0) is
Answers
Step-by-step explanation:
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Step-by-step explanation:
Let $A(0,0)$, $B(0,3)$, $C(4,0)$ be the given points. The excenter w.r.t. $A$ is $I_A(6,6)$ in the picture:
Excenter of the triangle with sides 3, 4, 5
First of all, it is on the angle bisector of the angle $\widehat{BAC}$, 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 $$ BC\ :\ \frac 35 x+\frac 45 y -\frac{12}5 =0\ , $$ 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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