1. **State the problem:** We have a square PQRS with diagonal PR. Points P and R are given as P(4,7) and R(8,-5). We need to find the equation of the line passing through points Q and S in the form $ay = bx + c$ where $a,b,c$ are integers. 2. **Find the midpoint M of diagonal PR:** Since PR is a diagonal of the square, the midpoint M of PR is also the midpoint of QS. $$M_x = \frac{4 + 8}{2} = 6, \quad M_y = \frac{7 + (-5)}{2} = 1$$ So, $M = (6,1)$. 3. **Find the vector PR:** $$\overrightarrow{PR} = (8 - 4, -5 - 7) = (4, -12)$$ 4. **Find the vector QS:** Since QS is the other diagonal of the square, it is perpendicular to PR and has the same length. The vector perpendicular to $\overrightarrow{PR} = (4, -12)$ is either $(12, 4)$ or $(-12, -4)$. 5. **Find coordinates of Q and S:** Let $\overrightarrow{QS} = (12, 4)$. Since M is midpoint of QS, $$Q = M + \frac{1}{2}\overrightarrow{QS} = (6 + 6, 1 + 2) = (12, 3)$$ $$S = M - \frac{1}{2}\overrightarrow{QS} = (6 - 6, 1 - 2) = (0, -1)$$ 6. **Find the equation of line QS:** Slope $m = \frac{3 - (-1)}{12 - 0} = \frac{4}{12} = \frac{1}{3}$. Using point-slope form with point Q(12,3): $$y - 3 = \frac{1}{3}(x - 12)$$ $$y - 3 = \frac{1}{3}x - 4$$ $$y = \frac{1}{3}x - 1$$ 7. **Rewrite in the form $ay = bx + c$ with integers:** Multiply both sides by 3: $$3y = x - 3$$ Rearranged: $$3y = 1x - 3$$ So, $a=3$, $b=1$, $c=-3$. **Final answer:** $$3y = x - 3$$