1963 AMC 12 Problem 33

Attempt Problem 33 of the 1963 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1963 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

33.

Given the line y=34x+6y=\dfrac34x+6 and a line LL parallel to the given line and 44 units from it. A possible equation for LL is:
y=34x+1y=\dfrac34x+1
y=34xy=\dfrac34x
y=34x−23y=\dfrac34x-\dfrac23
y=34x−1y=\dfrac34x-1
y=34x+2y=\dfrac34x+2
Answer: A
Concepts:parallel linesdistance formulaslope
Small hint:
Write parallel lines as 3x−4y+c=03x-4y+c=0
Big hint:
The distance equals the absolute difference of constants divided by 55
Solution:
The given line is 3x−4y+24=0.3x-4y+24=0. A parallel line y=3x4+by=\frac{3x}{4}+b is 3x−4y+4b=0,3x-4y+4b=0, so the distance is ∣24−4b∣5.\frac{|24-4b|}{5}. Setting this equal to 44 gives ∣6−b∣=5,|6-b|=5, hence b=1b=1 or 11.11. Choice A is possible.
Thus, the correct answer is A.
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