1960 AMC 12 Problem 30

Attempt Problem 30 of the 1960 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 1960 AMC 12 solutions, or check the answer key.

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30.

Given the line 3x+5y=153x+5y=15 and a point on this line equidistant from the coordinate axes. Such a point exists in:
none of the quadrants
quadrant I\mathrm{I} only
quadrants I,\mathrm{I}, II\mathrm{II} only
quadrants I,\mathrm{I}, II,\mathrm{II}, III\mathrm{III} only
each of the quadrants
Answer: C
Concepts:coordinate geometryabsolute valuesystem of equations
Small hint:
A point equidistant from the axes satisfies ∣x∣=∣y∣|x|=|y|
Big hint:
Intersect the given line with both y=xy=x and y=−xy=-x
Solution:
For y=x,y=x, the line gives 8x=15,8x=15, producing a point in quadrant I. For y=−x,y=-x, it gives −2x=15,-2x=15, so x<0x\lt0 and y>0,y>0, producing a point in quadrant II. There are no other possibilities.
Therefore, the correct answer is C.
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