SAT Question a35c7164: Linear Equations In Two Variables answer and explanation with Desmos solution

Choice B is correct. Two lines are perpendicular if their slopes are negative reciprocals, meaning that the slope of the first line is equal to $-1$ divided by the slope of the second line. Each equation in the given pair of equations can be written in slope-intercept form, $y=mx+b$, where $m$ is the slope of the graph of the equation in the xy-plane and $\left(0,b\right)$ is the y-intercept. For the first equation, $5x+7y=1$, subtracting $5x$ from both sides gives $7y=-5x+1$, and dividing both sides of this equation by $7$ gives $y=-\frac{5}{7}x+\frac{1}{7}$. Therefore, the slope of the graph of this equation is $-\frac{5}{7}$. For the second equation, $ax+by=1$, subtracting $ax$ from both sides gives $by=-ax+1$, and dividing both sides of this equation by $b$ gives $y=-\frac{a}{b}x+\frac{1}{b}$. Therefore, the slope of the graph of this equation is $-\frac{a}{b}$. Since the graph of the given pair of equations is a pair of perpendicular lines, the slope of the graph of the second equation, $-\frac{a}{b}$, must be the negative reciprocal of the slope of the graph of the first equation, $-\frac{5}{7}$. The negative reciprocal of $-\frac{5}{7}$ is $\frac{-1}{\left(-\frac{5}{7}\right)}$, or $\frac{7}{5}$. Therefore, $-\frac{a}{b}=\frac{7}{5}$, or $\frac{a}{b}=-\frac{7}{5}$. Similarly, rewriting the equations in choice B in slope-intercept form yields $y=-\frac{10}{7}x+\frac{1}{7}$ and $y=-\frac{a}{2b}x+\frac{1}{2b}$. It follows that the slope of the graph of the first equation in choice B is $-\frac{10}{7}$ and the slope of the graph of the second equation in choice B is $-\frac{a}{2b}$. Since $\frac{a}{b}=-\frac{7}{5}$, $-\frac{a}{2b}$ is equal to $-\left(\frac{1}{2}\right)\left(-\frac{7}{5}\right)$, or $\frac{7}{10}$. Since $\frac{7}{10}$ is the negative reciprocal of $-\frac{10}{7}$, the pair of equations in choice B represents a pair of perpendicular lines.