1. **Stating the problem:** Determine which among the sets A, B, or C have points that lie on the same straight line. 2. **Approach:** Points are collinear if the slope between any two pairs of points is the same. 3. **Check set A:** Points are $(2,3), (4,5), \left(-1\frac{a}{b}2,1\frac{a}{b}2\right)$. Assuming $-1\frac{a}{b}2 = -1 - \frac{a}{b} \times 2$ and $1\frac{a}{b}2 = 1 + \frac{a}{b} \times 2$, but since $a,b$ are unspecific variables, the slope cannot be confirmed. So set A is indeterminate without values for $a,b$. 4. **Check set B:** Points are $(2,3), (6,5), (1, \frac{1}{2})$. Calculate slope between $(2,3)$ and $(6,5)$: $$m_1 = \frac{5-3}{6-2} = \frac{2}{4} = \frac{1}{2}$$ Calculate slope between $(6,5)$ and $(1, \frac{1}{2})$: $$m_2 = \frac{\frac{1}{2} - 5}{1 - 6} = \frac{\frac{1}{2} - 5}{-5} = \frac{-\frac{9}{2}}{-5} = \frac{9}{10}$$ Since $m_1 = \frac{1}{2}$ and $m_2 = \frac{9}{10}$, slopes differ; thus points are not collinear. 5. **Check set C:** Points are $(2,4), (4,6), (-3, \frac{1}{2})$. Calculate slope between $(2,4)$ and $(4,6)$: $$m_1 = \frac{6-4}{4-2} = \frac{2}{2} = 1$$ Calculate slope between $(4,6)$ and $(-3, \frac{1}{2})$: $$m_2 = \frac{\frac{1}{2} - 6}{-3 - 4} = \frac{-\frac{11}{2}}{-7} = \frac{11}{14}$$ Slopes differ, so points are not collinear. 6. **Conclusion:** Without specific values of $a$ and $b$, set A cannot be confirmed. Sets B and C are not collinear. Final answer: **D) none**