1. **State the problem:** We are given a pair of linear equations $5x + 7y = 1$ and $ax + by = 1$ which represent perpendicular lines. We need to determine 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 the product of their slopes is $-1$. 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 $ax + by = 1$, rewrite as $y = -\frac{a}{b}x + \frac{1}{b}$. So slope $m_2 = -\frac{a}{b}$. 5. **Use perpendicularity condition:** $m_1 \times m_2 = -1$ implies $$\left(-\frac{5}{7}\right) \times \left(-\frac{a}{b}\right) = -1$$ Simplify: $$\frac{5a}{7b} = -1 \implies 5a = -7b \implies a = -\frac{7b}{5}$$ 6. **Check each option for perpendicularity:** - (a) Equations: $10x + 7y = 1$ and $ax - 2by = 1$ Slopes: $m_1 = -\frac{10}{7}$, $m_2 = -\frac{a}{-2b} = \frac{a}{2b}$ Product: $$m_1 m_2 = -\frac{10}{7} \times \frac{a}{2b} = -\frac{10a}{14b} = -\frac{5a}{7b}$$ Using $a = -\frac{7b}{5}$: $$-\frac{5 \times \left(-\frac{7b}{5}\right)}{7b} = -\frac{5 \times -7b/5}{7b} = -\frac{-7b}{7b} = 1 \neq -1$$ So not perpendicular. - (b) Equations: $10x + 7y = 1$ and $ax + 2by = 1$ Slopes: $m_1 = -\frac{10}{7}$, $m_2 = -\frac{a}{2b}$ Product: $$m_1 m_2 = -\frac{10}{7} \times -\frac{a}{2b} = \frac{10a}{14b} = \frac{5a}{7b}$$ Substitute $a = -\frac{7b}{5}$: $$\frac{5 \times -7b/5}{7b} = \frac{-7b}{7b} = -1$$ This satisfies perpendicularity. - (c) Equations: $10x + 7y = 1$ and $2ax + by = 1$ Slopes: $m_1 = -\frac{10}{7}$, $m_2 = -\frac{2a}{b}$ Product: $$m_1 m_2 = -\frac{10}{7} \times -\frac{2a}{b} = \frac{20a}{7b}$$ Substitute $a = -\frac{7b}{5}$: $$\frac{20 \times -7b/5}{7b} = \frac{-140b/5}{7b} = \frac{-28b}{7b} = -4 \neq -1$$ Not perpendicular. - (d) Equations: $5x - 7y = 1$ and $ax + by = 1$ Slopes: $m_1 = \frac{5}{7}$ (since $5x - 7y = 1 \implies y = \frac{5}{7}x - \frac{1}{7}$), $m_2 = -\frac{a}{b}$ Product: $$m_1 m_2 = \frac{5}{7} \times -\frac{a}{b} = -\frac{5a}{7b}$$ Substitute $a = -\frac{7b}{5}$: $$-\frac{5 \times -7b/5}{7b} = -\frac{-7b}{7b} = 1 \neq -1$$ Not perpendicular. 7. **Conclusion:** Only option (b) satisfies the perpendicularity condition. **Final answer:** The pair of equations in option (b) also represents a pair of perpendicular lines.