Line h is defined by 1 5 x + 1 7 y - 70 = 0

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 $y=mx+b$, 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 $-1$ 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 $\frac{-1}{\left(-{\displaystyle \frac{7}{5}}\right)}$, 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 $-\frac{7}{5}$.