Discussion

Find the coordinates of the incenter of the triangle whose vertices are (0, 0), (4, 0), and (0, 3).

Answer: A

Let the vertices be A(0,0), B(4,0), and C(0,3). First, find the lengths of the sides opposite these vertices.

Side a (BC) = \(\sqrt{(0-4)^2 + (3-0)^2} = \sqrt{16+9} = \sqrt{25} = 5\).

Side b (AC) = \(\sqrt{(0-0)^2 + (3-0)^2} = \sqrt{9} = 3\).

Side c (AB) = \(\sqrt{(4-0)^2 + (0-0)^2} = \sqrt{16} = 4\).

The formula for the incenter (I) is \(I = (\frac{ax_A+bx_B+cx_C}{a+b+c}, \frac{ay_A+by_B+cy_C}{a+b+c})\).

x-coordinate = \(\frac{5(0)+3(4)+4(0)}{5+3+4} = \frac{12}{12} = 1\).

y-coordinate = \(\frac{5(0)+3(0)+4(3)}{5+3+4} = \frac{12}{12} = 1\).

The incenter is at (1, 1).

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