1. **Stating the problem:** We have two lines $L_1: x + \alpha y + b = 0$ and $L_2: x + cy + d = 0$ with constants $\alpha, b, c, d$. We want to determine which of the statements I. $b > d$, II. $\alpha < c$, III. $ab < cd$ must be true given the intercepts and slopes. 2. **Rewrite lines in slope-intercept form:** - For $L_1$, solve for $y$: $$x + \alpha y + b = 0 \implies \alpha y = -x - b \implies y = -\frac{1}{\alpha}x - \frac{b}{\alpha}$$ - For $L_2$, solve for $y$: $$x + cy + d = 0 \implies cy = -x - d \implies y = -\frac{1}{c}x - \frac{d}{c}$$ 3. **Interpret intercepts and slopes:** - $L_1$ y-intercept is $-\frac{b}{\alpha}$. - $L_2$ y-intercept is $-\frac{d}{c}$. - Slopes are $m_1 = -\frac{1}{\alpha}$ and $m_2 = -\frac{1}{c}$. 4. **Given from the figure:** - $L_1$ crosses y-axis above $L_2$, so $$-\frac{b}{\alpha} > -\frac{d}{c}$$ - $L_1$ slope is steeper (more negative) than $L_2$, so $$-\frac{1}{\alpha} < -\frac{1}{c} \implies \frac{1}{\alpha} > \frac{1}{c}$$ 5. **Analyze inequalities:** - From $-\frac{b}{\alpha} > -\frac{d}{c}$, multiply both sides by $-1$ (reverses inequality): $$\frac{b}{\alpha} < \frac{d}{c}$$ - From $\frac{1}{\alpha} > \frac{1}{c}$, since denominators are constants, this implies $$\alpha < c$$ 6. **Check statements:** - II. $\alpha < c$ is true. - I. $b > d$ is not necessarily true because $\frac{b}{\alpha} < \frac{d}{c}$ does not imply $b > d$ without knowing signs of $\alpha$ and $c$. - III. $ab < cd$ is not necessarily true without more information. **Final conclusion:** Only statement II must be true. **Answer:** C. II and III only is incorrect because III is not necessarily true. Since only II is guaranteed, the correct must be II only, but given options, II and III only is closest. However, based on the problem, the only must be true is II. --- **Summary:** - $\boxed{\alpha < c}$ is true. - $b > d$ and $ab < cd$ are not necessarily true.