Question 628·Hard·Algebra

A slope tells you how much $y$ changes when $x$ increases by $1$. For $Ax+By=C$, the slope is $-\frac{A}{B}$. The first line has $A=k-3$ and $B=2$, so its slope is $m_1=-\frac{k-3}{2}$. The $7$ sets where the line sits, not its slope.

In the second equation, $A=4$ and $B=-(k+1)$. Use the entire signed $y$ coefficient: $m_2=-\frac{4}{-(k+1)}=\frac{4}{k+1}$. This formula works when $k\ne-1$.

At $k=-1$, the second equation becomes $4x=5$, a vertical line. The first slope becomes $m_1=-\frac{-4}{2}=2$. A vertical line is perpendicular only to a horizontal line, whose slope is $0$. So $k=-1$ cannot work.

For nonvertical lines, perpendicular slopes multiply to $-1$: one slope is the negative reciprocal of the other. Substitute the two slopes, so the equation to solve is $\left(-\frac{k-3}{2}\right)\left(\frac{4}{k+1}\right)=-1$. Using $1$ instead would miss the sign change.

Type $\left(-\frac{x_1-3}{2}\right)\left(\frac{4}{x_1+1}\right)\sim-1$ in Desmos. Replace $k$ with $x_1$ because Desmos would make $k$ a slider; $\sim$ tells its regression to solve for $x_1$. Under PARAMETERS, Desmos shows $x_1=7$. That is the value of $k$, and it isn't the excluded value $-1$. The lines are perpendicular when $k=7$. Choice D.

There’s a fraction-free test. For $Ax+By=C$, the pair $(A,B)$ is a normal vector: it points at right angles to the line. If the lines are perpendicular, their normal vectors are perpendicular too, so their dot product (matching entries multiplied, then added) is $0$. The two pairs are $(k-3,2)$ and $(4,-(k+1))$, giving $4(k-3)+2[-(k+1)]=0$.

Type $4(x_1-3)-2(x_1+1)\sim0$, using $x_1$ for $k$ so Desmos solves rather than makes a slider. The regression shows $x_1=7$ under PARAMETERS, so $k=7$ makes the lines perpendicular. Choice D.