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Algebra Difficulty: Hard

Line $h$ is defined by $\frac{1}{5} x + \frac{1}{7} y - 70 = 0$ . Line $j$ is perpendicular to line $h$ in the xy-plane. What is the slope of line $j$ ?

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Explanation

Choice D is correct. It’s given that line $h$ is defined by $\frac{1}{5} x + \frac{1}{7} y - 70 = 0$ . This equation can be written in slope-intercept form $y = m x + b$ , where $m$ is the slope of line $h$ and $b$ is the y-coordinate of the y-intercept of line $h$ . Adding $70$ to both sides of $\frac{1}{5} x + \frac{1}{7} y - 70 = 0$ yields $\frac{1}{5} x + \frac{1}{7} y = 70$ . Subtracting $\frac{1}{5} x$ from both sides of this equation yields $\frac{1}{7} y = - \frac{1}{5} x + 70$ . Multiplying both sides of this equation by $7$ yields $y = - \frac{7}{5} x + 490$ . Therefore, the slope of line $h$ is $- \frac{7}{5}$ . It’s given that line $j$ is perpendicular to line $h$ in the xy-plane. 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. Therefore, the slope of line $j$ is the negative reciprocal of the slope of line $h$ . The negative reciprocal of $- \frac{7}{5}$ is $\frac{- 1}{(- \frac{7}{5})}$ , or $\frac{5}{7}$ . Therefore, the slope of line $j$ is $\frac{5}{7}$ .

Choice A is incorrect. This is the slope of a line in the xy-plane that is parallel, not perpendicular, to line $h$ .

Choice B is incorrect. This is the reciprocal, not the negative reciprocal, of $- \frac{7}{5}$ .

Choice C is incorrect. This is the negative, not the negative reciprocal, of $- \frac{7}{5}$ .