1. **Problem statement:** We have a square with one vertex at $(1,2)$ and its diagonal lies along the line $$8x - 15y = 0.$$ We need to find the equations of the sides of the square passing through the vertex $(1,2)$. 2. **Understanding the problem:** The diagonal line is given by $$8x - 15y = 0,$$ which can be rewritten as $$y = \frac{8}{15}x.$$ The diagonal has slope $$m_d = \frac{8}{15}.$$ Since the diagonal passes through the center of the square, the sides of the square are perpendicular or parallel to the diagonal's direction. 3. **Key properties:** - The sides of the square are perpendicular to each other. - The diagonals of a square are perpendicular bisectors of each other. - The slope of the diagonal is $$m_d = \frac{8}{15}.$$ - The slope of the other diagonal $$m_{d2}$$ satisfies $$m_d \times m_{d2} = -1,$$ so $$m_{d2} = -\frac{15}{8}.$$ 4. **Finding slopes of sides:** The sides are perpendicular to the diagonals. Since the diagonals have slopes $$\frac{8}{15}$$ and $$-\frac{15}{8},$$ the sides have slopes that are perpendicular to these. - Slope perpendicular to $$\frac{8}{15}$$ is $$m_1 = -\frac{15}{8}.$$ - Slope perpendicular to $$-\frac{15}{8}$$ is $$m_2 = \frac{8}{15}.$$ 5. **Equations of sides passing through $(1,2)$:** Using point-slope form $$y - y_1 = m(x - x_1),$$ with point $(1,2)$: - Side 1: $$y - 2 = -\frac{15}{8}(x - 1)$$ - Side 2: $$y - 2 = \frac{8}{15}(x - 1)$$ 6. **Simplify equations:** - Side 1: $$y - 2 = -\frac{15}{8}x + \frac{15}{8}$$ $$y = -\frac{15}{8}x + \frac{15}{8} + 2 = -\frac{15}{8}x + \frac{15}{8} + \frac{16}{8} = -\frac{15}{8}x + \frac{31}{8}$$ - Side 2: $$y - 2 = \frac{8}{15}x - \frac{8}{15}$$ $$y = \frac{8}{15}x - \frac{8}{15} + 2 = \frac{8}{15}x - \frac{8}{15} + \frac{30}{15} = \frac{8}{15}x + \frac{22}{15}$$ **Final answer:** The equations of the sides of the square passing through $(1,2)$ are: $$y = -\frac{15}{8}x + \frac{31}{8}$$ and $$y = \frac{8}{15}x + \frac{22}{15}.$$