1. **State the problem:** We are given a line $L$ with equation $2x + 3y = 7$ and need to find which of the given lines is perpendicular to $L$. 2. **Recall the rule for perpendicular lines:** Two lines are perpendicular if the product of their slopes is $-1$. 3. **Find the slope of line $L$:** Rewrite $2x + 3y = 7$ in slope-intercept form $y = mx + b$. $$3y = -2x + 7$$ $$y = \frac{-2}{3}x + \frac{7}{3}$$ So, the slope of $L$ is $m = -\frac{2}{3}$. 4. **Check each option's slope and test perpendicularity:** - Option 1: $2x - 3y = 7$ $$-3y = -2x + 7$$ $$y = \frac{2}{3}x - \frac{7}{3}$$ Slope $= \frac{2}{3}$ Product with $L$'s slope: $$\left(-\frac{2}{3}\right) \times \frac{2}{3} = -\frac{4}{9} \neq -1$$ - Option 2: $3x + 2y = -7$ $$2y = -3x -7$$ $$y = -\frac{3}{2}x - \frac{7}{2}$$ Slope $= -\frac{3}{2}$ Product with $L$'s slope: $$\left(-\frac{2}{3}\right) \times \left(-\frac{3}{2}\right) = 1 \neq -1$$ - Option 3: $2x + 3y = -\frac{1}{7}$ This has the same left side as $L$, so slope is the same $-\frac{2}{3}$, not perpendicular. - Option 4: $3x - 2y = 7$ $$-2y = -3x + 7$$ $$y = \frac{3}{2}x - \frac{7}{2}$$ Slope $= \frac{3}{2}$ Product with $L$'s slope: $$\left(-\frac{2}{3}\right) \times \frac{3}{2} = -1$$ 5. **Conclusion:** Only option 4 has slope perpendicular to $L$. **Final answer:** The line $3x - 2y = 7$ is perpendicular to $2x + 3y = 7$.