1. **Problem statement:** Find the incenter (intersection of angle bisectors) and the inradius (radius of the incircle) of triangle ABC with vertices A(0|2), B(7|3), and C(10|6). Draw the incircle and give the coordinates of the incenter and the radius. 2. **Formula and rules:** - The incenter $W$ is the intersection of the three angle bisectors. - Coordinates of $W$ can be found using the formula for the incenter as a weighted average of vertices by side lengths: $$W = \left(\frac{a x_A + b x_B + c x_C}{a+b+c}, \frac{a y_A + b y_B + c y_C}{a+b+c}\right)$$ where $a = |BC|$, $b = |AC|$, $c = |AB|$ are the side lengths opposite vertices $A$, $B$, and $C$ respectively. - The inradius $r$ is the distance from $W$ to any side, calculated as the perpendicular distance. 3. **Calculate side lengths:** $$a = |BC| = \sqrt{(10-7)^2 + (6-3)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \approx 4.2426$$ $$b = |AC| = \sqrt{(10-0)^2 + (6-2)^2} = \sqrt{100 + 16} = \sqrt{116} \approx 10.7703$$ $$c = |AB| = \sqrt{(7-0)^2 + (3-2)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \approx 7.0711$$ 4. **Calculate coordinates of incenter $W$:** $$x_W = \frac{a x_A + b x_B + c x_C}{a+b+c} = \frac{4.2426 \cdot 0 + 10.7703 \cdot 7 + 7.0711 \cdot 10}{4.2426 + 10.7703 + 7.0711} = \frac{0 + 75.3921 + 70.711}{22.084} = \frac{146.1031}{22.084} \approx 6.615$$ $$y_W = \frac{a y_A + b y_B + c y_C}{a+b+c} = \frac{4.2426 \cdot 2 + 10.7703 \cdot 3 + 7.0711 \cdot 6}{22.084} = \frac{8.4852 + 32.3109 + 42.4266}{22.084} = \frac{83.2227}{22.084} \approx 3.769$$ 5. **Calculate inradius $r$:** Distance from $W$ to side $AB$ (line through A and B): - Equation of line $AB$: Slope $m = \frac{3-2}{7-0} = \frac{1}{7}$ Equation: $y - 2 = \frac{1}{7}(x - 0) \Rightarrow y = \frac{1}{7}x + 2$ - Distance from point $(x_W,y_W)$ to line $Ax + By + C = 0$ with $A = -\frac{1}{7}$, $B=1$, $C=-2$: $$r = \frac{|A x_W + B y_W + C|}{\sqrt{A^2 + B^2}} = \frac{\left| -\frac{1}{7} \cdot 6.615 + 3.769 - 2 \right|}{\sqrt{\left(-\frac{1}{7}\right)^2 + 1^2}} = \frac{| -0.945 + 3.769 - 2 |}{\sqrt{\frac{1}{49} + 1}} = \frac{|0.824|}{\sqrt{1.0204}} = \frac{0.824}{1.0102} \approx 0.816$$ 6. **Summary:** - Incenter $W \approx (6.615, 3.769)$ - Inradius $r \approx 0.816$ 7. **Drawing the incircle:** Draw triangle ABC with given points. Draw circle centered at $W$ with radius $r$ tangent to all sides. This completes the construction and calculation of the incircle for triangle ABC.