1. **Problem statement:** We have five vertices of a hexagon plotted on a coordinate plane: $A(1,5)$, $B(4,1)$, $C(9,1)$, $D(12,9)$, and $E(4,9)$. We need to find the coordinates of the sixth vertex $F$ such that all sides of the hexagon are equal in length. 2. **Key concept:** A regular hexagon has all sides equal. The distance between consecutive vertices is the same. We use the distance formula between two points $P(x_1,y_1)$ and $Q(x_2,y_2)$: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate side length:** Calculate the length of one known side, for example between $B(4,1)$ and $C(9,1)$: $$d = \sqrt{(9-4)^2 + (1-1)^2} = \sqrt{5^2 + 0} = 5$$ So each side length is 5 units. 4. **Check other sides:** Confirm other sides are length 5: - $AB$: $\sqrt{(4-1)^2 + (1-5)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5$ - $DE$: $\sqrt{(4-12)^2 + (9-9)^2} = \sqrt{(-8)^2 + 0} = 8$ (not 5, so $D$ and $E$ are not adjacent) 5. **Identify adjacent vertices:** The order of vertices around the hexagon must be such that each side is length 5. Given points, a possible order is $A-B-C-F-E-D-A$ or similar. 6. **Find $F$ coordinates:** Since $F$ is adjacent to $C(9,1)$ and $E(4,9)$, and $CF = EF = 5$, $F$ lies at the intersection of two circles: - Circle centered at $C(9,1)$ with radius 5: $$ (x-9)^2 + (y-1)^2 = 25 $$ - Circle centered at $E(4,9)$ with radius 5: $$ (x-4)^2 + (y-9)^2 = 25 $$ 7. **Solve system:** Subtract second equation from first: $$ (x-9)^2 + (y-1)^2 - (x-4)^2 - (y-9)^2 = 0 $$ Expand: $$ (x^2 - 18x + 81) + (y^2 - 2y + 1) - (x^2 - 8x + 16) - (y^2 - 18y + 81) = 0 $$ Simplify: $$ -18x + 81 - 2y + 1 + 8x - 16 + 18y - 81 = 0 $$ $$ (-18x + 8x) + (-2y + 18y) + (81 + 1 - 16 - 81) = 0 $$ $$ -10x + 16y - 15 = 0 $$ Rewrite: $$ 16y = 10x + 15 $$ $$ y = \frac{10x + 15}{16} $$ 8. **Substitute $y$ into one circle equation:** Use circle at $C$: $$ (x-9)^2 + \left(\frac{10x + 15}{16} - 1\right)^2 = 25 $$ Simplify inside the parenthesis: $$ \frac{10x + 15}{16} - 1 = \frac{10x + 15 - 16}{16} = \frac{10x - 1}{16} $$ So: $$ (x-9)^2 + \left(\frac{10x - 1}{16}\right)^2 = 25 $$ Multiply both sides by $16^2 = 256$ to clear denominator: $$ 256(x-9)^2 + (10x - 1)^2 = 25 \times 256 = 6400 $$ Expand: $$ 256(x^2 - 18x + 81) + (100x^2 - 20x + 1) = 6400 $$ $$ 256x^2 - 4608x + 20736 + 100x^2 - 20x + 1 = 6400 $$ $$ 356x^2 - 4628x + 20737 = 6400 $$ $$ 356x^2 - 4628x + 14337 = 0 $$ 9. **Solve quadratic:** Divide by 1 for simplicity: $$ 356x^2 - 4628x + 14337 = 0 $$ Use quadratic formula: $$ x = \frac{4628 \pm \sqrt{(-4628)^2 - 4 \times 356 \times 14337}}{2 \times 356} $$ Calculate discriminant: $$ 4628^2 = 21416884 $$ $$ 4 \times 356 \times 14337 = 20401032 $$ $$ \Delta = 21416884 - 20401032 = 1015842 $$ $$ \sqrt{1015842} \approx 1007.9 $$ So: $$ x = \frac{4628 \pm 1007.9}{712} $$ Two solutions: $$ x_1 = \frac{4628 + 1007.9}{712} \approx \frac{5635.9}{712} \approx 7.91 $$ $$ x_2 = \frac{4628 - 1007.9}{712} \approx \frac{3620.1}{712} \approx 5.08 $$ 10. **Find corresponding $y$ values:** $$ y_1 = \frac{10(7.91) + 15}{16} = \frac{79.1 + 15}{16} = \frac{94.1}{16} \approx 5.88 $$ $$ y_2 = \frac{10(5.08) + 15}{16} = \frac{50.8 + 15}{16} = \frac{65.8}{16} \approx 4.11 $$ 11. **Check which point fits hexagon shape:** The point $F(7.91, 5.88)$ fits better with the shape and spacing of the hexagon vertices. **Final answer:** One possible coordinate for the sixth vertex is approximately $$F(7.91, 5.88)$$