Question 448·Hard·Algebra

A line’s slope is its change in $y$ divided by its change in $x$. Type $\frac{7-(-3)}{10-4}$ in Desmos. It shows $1.666\ldots$; tap the fraction button to see $\frac{5}{3}$. Both subtractions go from $(4,-3)$ to $(10,7)$, so line $q$ has slope $\frac{5}{3}$.

Line $s$ is perpendicular to line $q$, so their slopes are negative reciprocals: flip $\frac{5}{3}$ and change its sign. Line $s$ has slope $-\frac{3}{5}$. Keeping $\frac{5}{3}$ would make the lines parallel, not perpendicular.

A midpoint sits halfway between the endpoints. Type $\frac{4+10}{2}$ and $\frac{-3+7}{2}$ on separate Desmos lines. They show $7$ and $2$, so the point on line $s$ is $(7,2)$. Average $x$ with $x$ and $y$ with $y$.

In $y=mx+b$, $m$ is the slope and $b$ is the $y$-intercept, the value of $y$ when $x=0$. Put $(7,2)$ and the slope $-\frac{3}{5}$ into that equation:

The midpoint is on line $s$, but its $y$-coordinate is not the intercept because its $x$-coordinate is not $0$.

Type $2\sim-\frac{3}{5}(7)+b$ in Desmos. The $\sim$ tells Desmos to find $b$; under PARAMETERS, it shows $b=6.2$. Type $b$ on the next line and tap its fraction button to see $\frac{31}{5}$. So the $y$-intercept of line $s$ is $\frac{31}{5}$. Choice A.