1. **State the problem:** We are given a triangle with vertices at points $(2, -6)$, $(5, 4)$, and $(k, 4)$, and the area of this triangle is 35 square units. We need to find the value of $k$. 2. **Formula for the area of a triangle given coordinates:** The area $A$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by: $$A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ 3. **Substitute the given points and area:** Let $(x_1, y_1) = (2, -6)$, $(x_2, y_2) = (5, 4)$, and $(x_3, y_3) = (k, 4)$. Area $A = 35$, so: $$35 = \frac{1}{2} \left| 2(4 - 4) + 5(4 - (-6)) + k(-6 - 4) \right|$$ 4. **Simplify inside the absolute value:** $$2(0) + 5(10) + k(-10) = 0 + 50 - 10k = 50 - 10k$$ 5. **Set up the equation:** $$35 = \frac{1}{2} |50 - 10k|$$ Multiply both sides by 2: $$70 = |50 - 10k|$$ 6. **Solve the absolute value equation:** $$50 - 10k = 70 \quad \text{or} \quad 50 - 10k = -70$$ For the first case: $$50 - 10k = 70$$ $$-10k = 20$$ $$k = -2$$ For the second case: $$50 - 10k = -70$$ $$-10k = -120$$ $$k = 12$$ 7. **Final answer:** The possible values of $k$ are $-2$ or $12$.