1. **Problem Statement:** We are given one line equation and points on a graph forming a quadrilateral. We need to find the equations of the other three lines connecting the points. 2. **Given:** - Line #1: $$\frac{x}{5} + \frac{y}{2} = 1$$ - Points on the graph: (-5,4), (8,-2), (-4,0), (5,0), (0,-4) - The plotted line passes through (-5,4) and (8,-2). 3. **Step 1: Confirm Line #1 equation in slope-intercept form.** Start with $$\frac{x}{5} + \frac{y}{2} = 1$$ Multiply both sides by 10 to clear denominators: $$10 \times \left(\frac{x}{5} + \frac{y}{2}\right) = 10 \times 1$$ $$2x + 5y = 10$$ Solve for $$y$$: $$5y = 10 - 2x$$ $$y = \frac{10 - 2x}{5} = 2 - \frac{2}{5}x$$ So, Line #1: $$y = 2 - \frac{2}{5}x$$ 4. **Step 2: Identify the vertices of the quadrilateral.** From the points and the line, the quadrilateral vertices are likely: - A = (-5,4) - B = (8,-2) - C = (5,0) - D = (-4,0) 5. **Step 3: Find equations of Lines #2, #3, and #4 connecting these points.** - Line #2 connects B(8,-2) and C(5,0): Slope $$m = \frac{0 - (-2)}{5 - 8} = \frac{2}{-3} = -\frac{2}{3}$$ Equation using point-slope form: $$y - (-2) = -\frac{2}{3}(x - 8)$$ $$y + 2 = -\frac{2}{3}x + \frac{16}{3}$$ $$y = -\frac{2}{3}x + \frac{16}{3} - 2 = -\frac{2}{3}x + \frac{16}{3} - \frac{6}{3} = -\frac{2}{3}x + \frac{10}{3}$$ - Line #3 connects C(5,0) and D(-4,0): Both points have $$y=0$$, so this is a horizontal line: $$y = 0$$ - Line #4 connects D(-4,0) and A(-5,4): Slope $$m = \frac{4 - 0}{-5 - (-4)} = \frac{4}{-1} = -4$$ Equation using point-slope form: $$y - 0 = -4(x + 4)$$ $$y = -4x - 16$$ 6. **Final answers:** - Line #2: $$y = -\frac{2}{3}x + \frac{10}{3}$$ - Line #3: $$y = 0$$ - Line #4: $$y = -4x - 16$$