1. **Problem statement:** Given two lines $L_1: 3x + ky - 2 = 0$ and $L_2: 6x + y + k = 0$, where $k$ is a non-zero constant, find the $y$-intercept of $L_2$ if the lines are perpendicular. 2. **Formula and rules:** - The slope of a line $Ax + By + C = 0$ is $m = -\frac{A}{B}$. - Two lines are perpendicular if the product of their slopes is $-1$, i.e., $m_1 \times m_2 = -1$. 3. **Find slopes:** - For $L_1$, slope $m_1 = -\frac{3}{k}$. - For $L_2$, slope $m_2 = -\frac{6}{1} = -6$. 4. **Apply perpendicularity condition:** $$m_1 \times m_2 = -1 \implies \left(-\frac{3}{k}\right) \times (-6) = -1$$ $$\frac{18}{k} = -1$$ 5. **Solve for $k$:** $$18 = -k \implies k = -18$$ 6. **Find $y$-intercept of $L_2$:** Rewrite $L_2$ as $y = -6x - k$. The $y$-intercept is the constant term when $x=0$, so: $$y\text{-intercept} = -k = -(-18) = 18$$ **Final answer:** The $y$-intercept of $L_2$ is $18$.