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I am teaching a student for SAT and I find the following problem. I have no idea what the notation $$ \fbox{k}=\left(-k,\frac{k}{2}\right)$$ means. Could you elaborate it more detailed?

The question reads:

$\fbox{k} = \left(-k, \frac{k}{2}\right)$ where $k$ is an integer. What is the equation of the line passing through $\fbox{k}$?

A. $y = 2x + 2$

B. $y = 2x$

C. $y = -2x$

D. $y = \frac{1}{2}x - 2$

E. $y = - \frac{1}{2}x$

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    $\begingroup$ I was seriously confused to find out this is not in any way related to the boolean satisfiability problem. $\endgroup$ – lisyarus Aug 3 '18 at 16:49
  • $\begingroup$ Is this an official CollegeBoard SAT question or one made from a third party test prep company? I've tutored SAT for years, and while they enjoy "weird notation" questions, this one doesn't quite feel like CollegeBoard's style. $\endgroup$ – zahbaz Aug 4 '18 at 6:50
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    $\begingroup$ "What is the equation of the line passing through [k]?" is not a logical question in this context. Because many lines can pass through the points represent by (−k,k/2). And in other to exclusively associate the line with the point, more data would be needed, for instance the gradient. $\endgroup$ – Igwe Kalu Aug 4 '18 at 10:53
  • $\begingroup$ @zahbaz: McGraw-Hill's Conquering SAT 2nd edition on page 264. $\endgroup$ – Field Medalist Aug 4 '18 at 13:59
  • $\begingroup$ @xport Okay, cool. McGraw-Hill usually has good practice problems. The actual SAT is often a bit more deliberate when introducing new notation. $\endgroup$ – zahbaz Aug 4 '18 at 17:11
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It appears that $\fbox{k}$ denotes a point (in this case, the point $(-k, k/2)$ for some integer $k$). It is not a notation I have ever seen before -- I would expect something like $P_k = (-k, k/2)$) -- but there is no accounting for taste.

(Although it wasn't in your question, the correct answer to the SAT question is then E: the line $y = -\frac{1}{2}x$ contains the point $ \fbox{k} = (-k, k/2)$ for every integer $k$.)

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    $\begingroup$ That it is new notation is the point. $\endgroup$ – Acccumulation Aug 3 '18 at 19:00
  • $\begingroup$ @Acccumulation, the new notation is a point. :) $\endgroup$ – Joel Reyes Noche Aug 4 '18 at 11:49
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You're always allowed to invent notation, as long as you explain what it means. Tests like the SAT like to do this to test if you're really comfortable with the ideas in abstract sense, and not tied to any particular notation.

But I have some objections to the way this is worded. The first sentence

$\fbox{k} = (-k, \frac{k}{2})$, where $k$ is an integer

seems to imply that $k$ is a single unknown value, so $\fbox{k}$ is a single point. But then there are an infinite number of lines through that point, so the question should be "which of these lines goes through that point?". The wording of the question seems to indicate that $\fbox{k}$ is a function defined on the integers, but then the question should say something like "... passing through all points $\fbox{k}$".

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    $\begingroup$ I have similar reservations about this question. "The wording of the question seems to indicate that $\fbox k$ is a function defined on the integers". Well, no, it doesn't suggest that $\fbox k$ is a function, it suggests that $\fbox{} $ is a function. If we had $f(k) = (-k,\frac k2)$, $f$ would be a function. $f(k)$ is not a function, it's a fixed value. I guess that "What is the equation of the set of points obtained by taking $\fbox k$ over all $k$?" would work, or "Which of these lines do we know for certain goes through $\fbox k$, without knowing what k is?" $\endgroup$ – Acccumulation Aug 3 '18 at 19:16
  • $\begingroup$ True, if k==0, Answers B,C,E work. If k=-1D,E works. $\endgroup$ – chux Aug 4 '18 at 5:29
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As MattPutnam explains in his answer, this question is asking the student to quickly adapt to a made-up notation.

The particular notation describes a "parametric equation":

$\fbox{k}$ defines an input parameter k, and an output point named $\fbox{k}$.

In this particular example:

  • The input is k, which can be any integer.
  • Thus, the student can choose from an infinite number of examples, and can even graph them.
  • The output (for any given value of k) is the point (-k, k/2).

You can think of the set of points {(-k, k/2) : k ϵ Z} as being a parametric equation of a set of (x, y) points along a line, where x and y have values determined by the parameter k.

To solve this question, set x = -k, and y = k/2. Either check each proposed answer by substitution (CBS), or solve for y in terms of x:

x = -k.
k = -x.
y = -x/2.
Answer E is correct.
CBS:
y   ≟ -x/2
k/2 ≟ -(-k)/2
k/2 =  k/2
Confirmed that Answer E is correct.

Alternatively, the student could choose a value of k, evaluate $\fbox{k}$, perform a CBS for each answer, and repeat until only one answer passed the CBS. For example:

CBS:
Suppose k = 0.
$\fbox{k}$ = (0, 0)
A is wrong: (0, 2) ≠ (0, 0).
D is wrong: (0, -2) ≠ (0, 0).
B, C, and E are still possible: (0, 0) = (0, 0).

Suppose k = 2.
$\fbox{k}$ = (-2, 1)
B is wrong: 1 ≠ 2*(-2).
C is wrong: 1 ≠ -2*(-2).
E is OK: 1 = (-1/2)*(-2).

Therefore Answer E is correct.

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