Question #a35c7164 - SAT Suite Question Bank

Modded SAT Question Bank by Abdullah Mallik

$5 x + 7 y = 1$

$a x + b y = 1$

In the given pair of equations, $a$ and $b$ are constants. The graph of this pair of equations in the xy-plane is a pair of perpendicular lines. Which of the following pairs of equations also represents a pair of perpendicular lines?

$10 x + 7 y = 1$

$a x - 2 b y = 1$
$10 x + 7 y = 1$

$a x + 2 b y = 1$
$10 x + 7 y = 1$

$2 a x + b y = 1$
$5 x - 7 y = 1$

$a x + b y = 1$

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Correct Answer: B

Rationale

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 $negative 1$ divided by the slope of the second line. Each equation in the given pair of equations can be written in slope-intercept form, $y = m x + b$ , where $m$ is the slope of the graph of the equation in the xy-plane and $left parenthesis 0 comma b right parenthesis$ is the y-intercept. For the first equation, $5 x + 7 y = 1$ , subtracting $5 x$ from both sides gives $7 y = - 5 x + 1$ , 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, $a x + b y = 1$ , subtracting $a x$ from both sides gives $b y = - a x + 1$ , 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 $\frac{- 1}{(- \frac{5}{7})}$ , 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 $y = - \frac{a}{2 b} x + \frac{1}{2 b}$ . 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 $- \frac{a}{2 b}$ . Since $\frac{a}{b} = - \frac{7}{5}$ , $- \frac{a}{2 b}$ is equal to $- (\frac{1}{2}) (- \frac{7}{5})$ , 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.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

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