1. **State the problem:** We have three lines L₁, L₂, and L₃ with given intercepts and relationships: L₁ has x-intercept -6 and y-intercept 4, L₁ is perpendicular to L₂, and L₁ is parallel to L₃. We need to verify which statements I, II, and III are true. 2. **Find the equation and slope of L₁:** - The x-intercept is (-6, 0) and y-intercept is (0, 4). - The slope $m_1$ of L₁ is given by $$m_1 = \frac{4 - 0}{0 - (-6)} = \frac{4}{6} = \frac{2}{3}.$$ - Equation of L₁ using intercept form: $$\frac{x}{-6} + \frac{y}{4} = 1.$$ Or slope-intercept form: $$y = \frac{2}{3}x + 4.$$ 3. **Check statement I: Does L₁ pass through (8, 6)?** - Substitute $x=8$ into L₁: $$y = \frac{2}{3} \times 8 + 4 = \frac{16}{3} + 4 = \frac{16}{3} + \frac{12}{3} = \frac{28}{3} \approx 9.33.$$ - The point (8, 6) has $y=6$, which is not equal to 9.33, so **statement I is false**. 4. **Find the slope of L₂:** - Since L₁ is perpendicular to L₂, their slopes satisfy: $$m_1 \times m_2 = -1.$$ - We have $m_1 = \frac{2}{3}$, so $$m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}.$$ - This matches statement II, so **statement II is true**. 5. **Find the equation of L₃:** - L₃ is parallel to L₁, so $m_3 = m_1 = \frac{2}{3}$. - L₃ passes through (9, 0), so use point-slope form: $$y - 0 = \frac{2}{3}(x - 9) \Rightarrow y = \frac{2}{3}x - 6.$$ - Rearranged to standard form: $$y = \frac{2}{3}x - 6 \Rightarrow 3y = 2x - 18 \Rightarrow 2x - 3y - 18 = 0.$$ - Statement III claims the equation is $6x - 9y - 16 = 0$, which is different from $2x - 3y - 18 = 0$, so **statement III is false**. **Final conclusion:** Only statement II is true. **Answer: B. II only**