1. **State the problem:** We have a triangle ABC with vertices A(-2,1), B(6,5), and C(4,k). Angle ABC is the right angle. We need to find the equation of the line passing through points A and C in the form $ay + bx = c$ where $a,b,c$ are integers. 2. **Use the right angle condition:** Since angle ABC is right, vectors BA and BC are perpendicular. 3. **Find vectors BA and BC:** $$\vec{BA} = (x_A - x_B, y_A - y_B) = (-2 - 6, 1 - 5) = (-8, -4)$$ $$\vec{BC} = (x_C - x_B, y_C - y_B) = (4 - 6, k - 5) = (-2, k - 5)$$ 4. **Use the dot product for perpendicularity:** $$\vec{BA} \cdot \vec{BC} = 0$$ $$(-8)(-2) + (-4)(k - 5) = 0$$ $$16 - 4(k - 5) = 0$$ 5. **Solve for $k$:** $$16 - 4k + 20 = 0$$ $$36 - 4k = 0$$ $$-4k = -36$$ $$k = 9$$ 6. **Now find the equation of line through A(-2,1) and C(4,9):** Slope $m$: $$m = \frac{9 - 1}{4 - (-2)} = \frac{8}{6} = \frac{4}{3}$$ 7. **Use point-slope form:** $$y - y_1 = m(x - x_1)$$ $$y - 1 = \frac{4}{3}(x + 2)$$ 8. **Expand and simplify:** $$y - 1 = \frac{4}{3}x + \frac{8}{3}$$ $$y = \frac{4}{3}x + \frac{8}{3} + 1$$ $$y = \frac{4}{3}x + \frac{11}{3}$$ 9. **Convert to form $ay + bx = c$ with integers:** Multiply both sides by 3: $$3y = 4x + 11$$ Rewrite: $$3y - 4x = 11$$ This is the required equation. **Final answer:** $$3y - 4x = 11$$