1. **State the problem:** We have a pair of linear equations $5x + 7y = 1$ and $ax + by = 1$ that represent perpendicular lines. We want to find which of the given pairs of equations also represent perpendicular lines. 2. **Recall the condition for perpendicular lines:** Two lines $A_1x + B_1y = C_1$ and $A_2x + B_2y = C_2$ are perpendicular if their slopes satisfy $m_1 \cdot m_2 = -1$. 3. **Find the slope of the first line:** For $5x + 7y = 1$, rewrite as $y = -\frac{5}{7}x + \frac{1}{7}$, so slope $m_1 = -\frac{5}{7}$. 4. **Find the slope of the second line:** For $ax + by = 1$, slope $m_2 = -\frac{a}{b}$. 5. **Use the perpendicular condition:** $$m_1 \cdot m_2 = -1 \implies \left(-\frac{5}{7}\right) \cdot \left(-\frac{a}{b}\right) = -1$$ Simplify: $$\frac{5a}{7b} = -1 \implies 5a = -7b \implies a = -\frac{7b}{5}$$ 6. **Check each option for perpendicularity:** - Option A: Equations: $10x + 7y = 1$ and $ax - 2by = 1$ Slopes: $m_1 = -\frac{10}{7}$, $m_2 = -\frac{a}{-2b} = \frac{a}{2b}$ Perpendicular condition: $$m_1 \cdot m_2 = -1 \implies -\frac{10}{7} \cdot \frac{a}{2b} = -1 \implies -\frac{10a}{14b} = -1 \implies \frac{10a}{14b} = 1 \implies 5a = 7b \implies a = \frac{7b}{5}$$ This contradicts $a = -\frac{7b}{5}$, so not perpendicular. - Option B: Equations: $10x + 7y = 1$ and $ax + 2by = 1$ Slopes: $m_1 = -\frac{10}{7}$, $m_2 = -\frac{a}{2b}$ Perpendicular condition: $$-\frac{10}{7} \cdot \left(-\frac{a}{2b}\right) = -1 \implies \frac{10a}{14b} = -1 \implies 5a = -7b \implies a = -\frac{7b}{5}$$ This matches the original condition, so option B represents perpendicular lines. - Option C: Equations: $10x + 7y = 1$ and $2ax + by = 1$ Slopes: $m_1 = -\frac{10}{7}$, $m_2 = -\frac{2a}{b}$ Perpendicular condition: $$-\frac{10}{7} \cdot \left(-\frac{2a}{b}\right) = -1 \implies \frac{20a}{7b} = -1 \implies 20a = -7b \implies a = -\frac{7b}{20}$$ This does not match $a = -\frac{7b}{5}$, so not perpendicular. - Option D: Equations: $5x - 7y = 1$ and $ax + by = 1$ Slopes: $m_1 = \frac{5}{7}$, $m_2 = -\frac{a}{b}$ Perpendicular condition: $$\frac{5}{7} \cdot \left(-\frac{a}{b}\right) = -1 \implies -\frac{5a}{7b} = -1 \implies \frac{5a}{7b} = 1 \implies 5a = 7b \implies a = \frac{7b}{5}$$ This contradicts $a = -\frac{7b}{5}$, so not perpendicular. **Final answer:** Only option B represents a pair of perpendicular lines.