Aniko

Parameterized points and answer choices

Lesson progressPractice problems 0/3
Difficulty
Advanced
Estimated time
25 minutes
Domains
Algebra
Techniques
ParameterizationAnswer-choice-testingParametric-pathsPath-classification

What you’ll learn

  1. Recognize coincident lines and parameter-in-choice wording on the SAT.
  2. Turn parameterized answer choices into paths with Desmos’s variable t.
  3. Interpret overlap, one intersection, and parallel paths as every value, exactly one value, and no values.
  4. Confirm an every answer with a substitution identity.

Prerequisites

Why this matters on the SAT

Match each path to the shared line

Some SAT systems represent the same line, so they have infinitely many solutions. Instead of asking for one intersection, the test may ask which formula gives a point on both lines for every value of a parameter.

The answer choices can look algebra-heavy, but Desmos turns them into paths. Graph the system, compare the paths, and find the one that stays on the shared line. This is fast and makes sign or coordinate-order mistakes easy to spot.

SAT example

The system of equations is

8x−6y=108x-6y=10

and

−12x+9y=−15.-12x+9y=-15.

For each real number rr, which point lies on the graph of each equation in the xyxy-plane?

  1. A

    (r,5−4r3)\left(r,\frac{5-4r}{3}\right)

  2. B

    (4r−53,r)\left(\frac{4r-5}{3},r\right)

  3. C

    (r,4r−53)\left(r,\frac{4r-5}{3}\right)

  4. D

    (r,4r+53)\left(r,\frac{4r+5}{3}\right)

Fast Desmos solution

Enter both equations. They overlap, so the system represents one line. Replace rr with Desmos's path variable tt in the choices. Only choice C, (t,(4t-5)/3), traces the shared line, so the answer is C.

That visual match is the core skill in this lesson. Next, you will learn how to build the paths, interpret their relationships, and confirm the answer.

Calculator loads as you approach
Choice C traces the same path as the system.

1. Turn a parameterized point into a path

A parameter is a variable allowed to take any permitted value. An answer choice such as

(r,3r−52)\left(r,\frac{3r-5}{2}\right)

does not name one fixed point. It describes a moving point: x=rx=r, and the second coordinate gives the matching yy-value.

If you enter the choice with rr, Desmos creates a slider and displays one point at a time. Moving the slider builds intuition and can disprove a choice as soon as the point leaves the target line.

To see a continuous path segment at once, replace the problem's parameter with Desmos's special variable tt:

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

Desmos initially shows a limited tt-interval. Extend it to −10≤t≤10-10\le t\le10 when you need a wider comparison. Keep the coordinates in their original order when you make the replacement.

Calculator loads as you approach
The r-slider controls one blue point. Desmos t reveals the green path segment.

2. Read every, exactly one, and never from the graph

When the parameter moves each candidate point along a line, as it does in this lesson, Desmos turns the wording into three visible relationships:

  • Every value: the candidate path overlaps the target line.
  • Exactly one value: the candidate path crosses the target line without overlapping it.
  • No values: the candidate path does not touch the target line. In these questions, it is often parallel.

The graph gives the diagnosis. Substitution explains why:

  • overlap produces an identity such as 7=77=7;
  • one intersection produces an equation such as 4r=84r=8;
  • no intersection produces a contradiction such as 5=75=7.
Check your understanding:

Use the graph to compare P(r)=(r,2r−1)P(r)=(r,2r-1), Q(r)=(r,r+1)Q(r)=(r,r+1), and R(r)=(r,2r+3)R(r)=(r,2r+3) with y=2x−1y=2x-1. Which works for every real rr, which for exactly one value of rr, and which for no values?

The three path relationships in the calculator match the three algebraic outcomes.

Common mistake:

Treating finite graph evidence as proof about all real parameter values. One failed value disproves "every," but successful samples cannot prove every value. A finite visible path also cannot prove "no values." Before concluding no values, use substitution to produce a contradiction or confirm algebraically that the paths are distinct parallel lines.

Calculator loads as you approach
The green path overlaps, the blue path crosses once, and the red path stays parallel.

3. Example: Compare the paths, then confirm

Worked example

The system of equations is

6x−4y=106x-4y=10

and

−3x+2y=−5.-3x+2y=-5.

For every real number rr, which point lies on the graph of each equation?

  1. A

    (r,3r−52)\left(r,\frac{3r-5}{2}\right)

  2. B

    (r,3r+52)\left(r,\frac{3r+5}{2}\right)

  3. C

    (3r−52,r)\left(\frac{3r-5}{2},r\right)

  4. D

    (r,5−3r2)\left(r,\frac{5-3r}{2}\right)

Step 1

Graph the system

Enter both equations exactly as written. Their graphs overlap, so the system has infinitely many solutions and can be treated as one shared line.

As a quick check, multiplying the second equation by −2-2 gives 6x−4y=106x-4y=10.

Step 2

Turn the choices into paths

Replace rr with Desmos's path variable tt in each ordered pair:

  • A: (t,(3t-5)/2)
  • B: (t,(3t+5)/2)
  • C: ((3t-5)/2,t)
  • D: (t,(5-3t)/2)

Extend each path to −10≤t≤10-10\le t\le10 so the relationships are easy to compare.

Check your understanding:

Before inspecting the graph, predict whether choice B, (t,(3t+5)/2), will overlap, cross, or stay parallel to the target.

Calculator loads as you approach
Choice A overlaps the shared line. The other paths are parallel or cross it.

Step 3

Read the graph

Choice A is the only path that overlaps the shared line. Choice B stays parallel. Choices C and D each cross the target once, so they work for exactly one parameter value, not every value.

The graph identifies A quickly and exposes the distractors: B changes a sign, C swaps the coordinates, and D reverses the expression.

Step 4

Confirm the winner exactly

Substitute choice A into the shared line:

6r−4(3r−52)=6r−(6r−10)=10.6r-4\left(\frac{3r-5}{2}\right)=6r-(6r-10)=10.

The identity 10=1010=10 proves that choice A works for every real rr. Because the equations are equivalent, the point lies on both.

The Desmos path finds the answer. The identity confirms that the visible overlap continues for every real parameter value.

4. Use algebra when it is obviously shorter

Desmos is the default method for this lesson, but a short algebraic pattern can still save time.

Use Desmos first when…

  • several parameterized choices look alike;
  • both coordinates depend on the parameter; or
  • the wording asks you to distinguish every value, exactly one value, and no values.

Use the algebra shortcut when…

  • one coordinate already equals the parameter;
  • isolating the other coordinate takes one or two clear steps; and
  • you can preserve the ordered-pair position without extra work.

For 5x+2y=75x+2y=7 with choices of the form (r,□)(r,\square), set x=rx=r to get y=7−5r2y=\frac{7-5r}{2}. The point is (r,7−5r2)\left(r,\frac{7-5r}{2}\right).

Finish the solution

The shared line and choice A are already in the calculator. Enter the remaining choices with tt in place of rr, find the overlapping path, and confirm the coordinate order.

Complete the reversed-coordinate formula

Finish the solution

The system of equations is

4(2x−3y)+5=116y−4x=−3.\begin{aligned} 4(2x-3y)+5 &= 11 \\ 6y-4x &= -3. \end{aligned}

For each real number rr, which point lies on the graph of each equation?

First steps

  1. Simplify both equations to 4x−6y=34x-6y=3.
  2. The shared line and choice A's path are already entered.
  3. Replace rr with tt in choices B through D and enter those paths.

Finish it

Answer choices
Calculator loads as you approach
Choice A is entered. Add choices B through D and find the path that overlaps the shared line.

Practice problems

SAT practice problems

Graph the candidate paths first. Use substitution to confirm why your answer works.

Find the never-on-line path

Practice problem

The target line is

3x−5y=11.3x-5y=11.

Which parameterized point never lies on the line for any real value of rr?

Answer choices
Calculator loads as you approach
Enter the target line and each candidate with t, then widen every path's parameter domain from -10 to 10. The no-values path should not touch the target.

Compare paths with two moving coordinates

Practice problem

The system of equations is

3(2x+y)−4=113(2x+y)-4=11

and

4x+2y=10.4x+2y=10.

For every real number rr, which point lies on the graph of both equations?

Answer choices
Calculator loads as you approach
Graph the shared line and candidate paths with Desmos t. Then prove which path overlaps for every value.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

  • A parameterized ordered pair describes a moving point.
  • Replace the problem's parameter with Desmos's variable t to display a continuous candidate path segment.
  • When the parameter moves a point along a line, overlap means every value, one intersection means exactly one value, and no intersection means no values.
  • Preserve coordinate order when you enter a path or use the algebra shortcut.
  • An identity confirms that an overlapping path works for every permitted value. One failed value disproves "every," but successful samples do not prove it.

Next lesson

Factor, root, and remainder tests

Use strategic substitutions and shared roots to test algebraic answer choices.

Start next lesson
Parameterized Points in Desmos for the SAT | aniko.ai