1. **Problem:** Find the value of $a$ such that the lines $5x + 3y - 4 = 0$ and $ax + 3y - 5 = 0$ are perpendicular. 2. **Formula and rule:** Two lines $Ax + By + C = 0$ and $A'x + B'y + C' = 0$ are perpendicular if and only if their slopes satisfy $m_1 \cdot m_2 = -1$. The slope of a line $Ax + By + C = 0$ is $m = -\frac{A}{B}$ (assuming $B \neq 0$). 3. **Calculate slopes:** - For the first line: $m_1 = -\frac{5}{3}$ - For the second line: $m_2 = -\frac{a}{3}$ 4. **Apply perpendicularity condition:** $$m_1 \cdot m_2 = -1 \implies \left(-\frac{5}{3}\right) \cdot \left(-\frac{a}{3}\right) = -1$$ 5. **Simplify:** $$\frac{5a}{9} = -1$$ 6. **Solve for $a$:** $$a = -\frac{9}{5}$$ **Final answer:** $a = -\frac{9}{5}$