Parameterized points and answer choices

Some SAT systems are really one line written two ways, so they have infinitely many solutions. There's no single intersection to ask for. So the question asks something else: which formula gives a point on both lines for every value of $r$?

The answer choices look heavy with algebra, but Desmos turns each one into a path you can see. Graph the system, graph the choices, and pick the one that stays on the shared line. It's fast, and a flipped sign or a swapped pair of coordinates shows up right away.

Here's one.

Fast Desmos solution

Type both equations. They land right on top of each other, so the system is really one line. Now type the choices with $t$ in place of $r$, because $t$ is the letter Desmos uses to draw a path. Only choice C, (t,(4t-5)/3), runs along the shared line, so the answer is C.

A parameter is a letter that can take any allowed value. So an answer choice like

isn't one fixed point. It's a point that moves as $r$ changes. Its $x$-coordinate is $r$, and its $y$-coordinate is $\frac{3r-5}{2}$. At $r=2$, it's the point $\left(2,\frac12\right)$. At $r=3$, it's $(3,2)$.

Desmos keeps $r$ for polar graphs, so type the choice with another letter, like (a,(3a-5)/2). Desmos makes a slider for $a$ and shows one point at a time. Drag the slider and watch the point move. That helps you get a feel for it, and if the point ever leaves the target line, you can cross that choice off.

To see a whole stretch of the path at once, swap the problem's letter for $t$. Desmos treats $t$ as a special path variable:

(t,(3t-5)/2)

At first, Desmos draws only a short piece of the path. When you need more of it to compare, widen the range to $-10\le t\le10$.

One thing to watch: swap only the letter, and keep the coordinates in the order the choice gives them. A choice with the coordinates swapped draws a different path, and that's a favorite wrong answer.

In these questions, each choice's point moves along a straight line as the parameter changes. So its path can meet the target line in only three ways, and each one answers the question differently:

• If the path lies right on top of the target line, the choice works for every value.
• If the path crosses the target line once, the choice works for exactly one value, the one at the crossing.
• If the path never touches the target line, the choice works for no values. Here that often means the path runs parallel to it.

The graph tells you which case you're in. Substitution tells you why. Plug the choice into the target equation and look at what's left:

• Overlap leaves an identity, a statement that's always true, like $7=7$.
• One crossing leaves an equation with one solution, like $4r=8$.
• No crossing leaves a contradiction, a statement that's never true, like $5=7$.