1. **State the problem:** We are given a line $L$ with equation $3y - 4x = 21$ and a point $P(15, 2)$. We need to find the equation of the line perpendicular to $L$ that passes through $P$. 2. **Rewrite the equation of line $L$ in slope-intercept form:** $$3y - 4x = 21$$ Add $4x$ to both sides: $$3y = 4x + 21$$ Divide both sides by 3: $$y = \frac{\cancel{3}4}{\cancel{3}3}x + \frac{21}{3}$$ Simplify: $$y = \frac{4}{3}x + 7$$ 3. **Find the slope of line $L$:** The slope $m_L = \frac{4}{3}$. 4. **Find the slope of the perpendicular line:** The slope of a line perpendicular to another is the negative reciprocal of the original slope. $$m_\perp = -\frac{1}{m_L} = -\frac{1}{\frac{4}{3}} = -\frac{3}{4}$$ 5. **Use point-slope form to find the equation of the perpendicular line passing through $P(15, 2)$:** $$y - y_1 = m_\perp (x - x_1)$$ Substitute $m_\perp = -\frac{3}{4}$, $x_1 = 15$, and $y_1 = 2$: $$y - 2 = -\frac{3}{4}(x - 15)$$ 6. **Simplify the equation:** $$y - 2 = -\frac{3}{4}x + \frac{3}{4} \times 15$$ Calculate $\frac{3}{4} \times 15$: $$\frac{3}{4} \times 15 = \frac{3 \times 15}{4} = \frac{45}{4}$$ So: $$y - 2 = -\frac{3}{4}x + \frac{45}{4}$$ Add 2 to both sides: $$y = -\frac{3}{4}x + \frac{45}{4} + 2$$ Convert 2 to quarters: $$2 = \frac{8}{4}$$ So: $$y = -\frac{3}{4}x + \frac{45}{4} + \frac{8}{4} = -\frac{3}{4}x + \frac{53}{4}$$ **Final answer:** $$\boxed{y = -\frac{3}{4}x + \frac{53}{4}}$$