1. **Problem statement:** Find the value of $k$ such that the area of the triangle with vertices $(2, -6)$, $(5, 4)$, and $(k, 4)$ is 35 square units. 2. **Formula for area of triangle given vertices:** $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ 3. **Assign vertices:** $$ (x_1, y_1) = (2, -6), \quad (x_2, y_2) = (5, 4), \quad (x_3, y_3) = (k, 4) $$ 4. **Substitute into formula:** $$\text{Area} = \frac{1}{2} |2(4 - 4) + 5(4 + 6) + k(-6 - 4)|$$ 5. **Simplify inside absolute value:** $$2(0) + 5(10) + k(-10) = 0 + 50 - 10k = 50 - 10k$$ 6. **Set area equal to 35:** $$35 = \frac{1}{2} |50 - 10k| \implies 70 = |50 - 10k|$$ 7. **Solve absolute value equation:** $$50 - 10k = 70 \quad \text{or} \quad 50 - 10k = -70$$ 8. **First case:** $$50 - 10k = 70 \implies -10k = 20 \implies k = -2$$ 9. **Second case:** $$50 - 10k = -70 \implies -10k = -120 \implies k = 12$$ 10. **Final answer:** $$k = -2 \quad \text{or} \quad k = 12$$