Find the incenter of the triangle formed by t | vedclass

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(B) The vertices of the triangle formed by the axes and the line $\frac{x}{a} + \frac{y}{b} = 1$ are $O(0, 0)$,$A(a, 0)$,and $B(0, b)$.Let the side le

Vedclass · Accepted Answer

$\left( \frac{ab}{a + b + \sqrt{a^2 + b^2}}, \frac{ab}{a + b + \sqrt{a^2 + b^2}} \right)$

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(B) The vertices of the triangle formed by the axes and the line $\frac{x}{a} + \frac{y}{b} = 1$ are $O(0, 0)$,$A(a, 0)$,and $B(0, b)$.Let the side lengths be $c = AB = \sqrt{a^2 + b^2}$,$b' = OB = b$,and $a' = OA = a$.The coordinates of the incenter $(I_x, I_y)$ are given by the formula $\left( \frac{a'x_1 + b'x_2 + c'x_3}{a' + b' + c'}, \frac{a'y_1 + b'y_2 + c'y_3}{a' + b' + c'} \right)$.Substituting the vertices $(0, 0)$,$(a, 0)$,and $(0, b)$ and side lengths $a' = b$,$b' = a$,and $c' = \sqrt{a^2 + b^2}$:$I_x = \frac{b(0) + a(a) + \sqrt{a^2 + b^2}(0)}{a + b + \sqrt{a^2 + b^2}} = \frac{ab}{a + b + \sqrt{a^2 + b^2}}$$I_y = \frac{b(0) + a(0) + \sqrt{a^2 + b^2}(b)}{a + b + \sqrt{a^2 + b^2}} = \frac{ab}{a + b + \sqrt{a^2 + b^2}}$Thus,the incenter is $\left( \frac{ab}{a + b + \sqrt{a^2 + b^2}}, \frac{ab}{a + b + \sqrt{a^2 + b^2}} \right)$.

Find the incenter of the triangle formed by the axes and the line $\frac{x}{a} + \frac{y}{b} = 1$.

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