Question 462·Hard·Algebra

Line $\ell$ has $y$ by itself. In this form, the slope is the coefficient of $x$, so its slope is $k-1$. The $2k$ is the $y$-intercept, not the slope.

From $(3,4)$ to $(k,10)$, put the change in $y$ over the change in $x$, keeping the subtractions in the same order:

Here $k\ne3$. Otherwise $m$ would be vertical, while $\ell$ would have slope $2$, not the horizontal slope $0$ needed to be perpendicular to it.

The lines are perpendicular, meaning they meet at a right angle. For these nonvertical lines, their slopes multiply to $-1$, so:

Because $k\ne3$, multiply both sides by $k-3$ to get an equation without a fraction:

In Desmos, type $6(x_1-1)\sim-(x_1-3)$. Use $x_1$ for the unknown $k$ and $\sim$ to tell Desmos to fit a value. Under PARAMETERS, it shows $x_1\approx1.285714$. That’s the value of $k$ as a decimal.

Add $x_1$ on its own Desmos line and tap its fraction button. Desmos shows the exact fraction $\frac{9}{7}$. Since $x_1$ stands for $k$, the value of $k$ is $\frac{9}{7}$. Choice B.

Start with the equation from the perpendicular slopes. Distribute through each set of parentheses:

Add $k$ to both sides of this linear equation:

Add $6$ to both sides, so only the $k$ term remains on the left:

Divide both sides by $7$:

The coefficient of $k$ was $7$, so divide $9$ by $7$, not the other way around. Choice B.