1960 AHSME Problems/Problem 30

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Problem

Given the line $3x+5y=15$ and a point on this line equidistant from the coordinate axes. Such a point exists in:

$\textbf{(A)}\ \text{none of the quadrants}\qquad\textbf{(B)}\ \text{quadrant I only}\qquad\textbf{(C)}\ \text{quadrants I, II only}\qquad$ $\textbf{(D)}\ \text{quadrants I, II, III only}\qquad\textbf{(E)}\ \text{each of the quadrants}$

Solution

If a point is equidistant from the coordinate axes, then the absolute values of the x-coordinate and y-coordinate are equal. Since the point is on the line $3x+5y=15$, find the intersection point of $y=x$ and $3x+5y=15$ and the intersection point of $y=-x$ and $3x+5y=15$.

Substituting $x$ for $y$ in $3x+5y=15$ results in $8x=15$, so $x=\frac{15}{8}$. That means $y=\frac{15}{8}$, so one of the points is in the first quadrant.

Substituting $-x$ and $y$ in $3x+5y=15$ results in $-2x=15$, so $x=\frac{-15}{2}$. That means $y=\frac{15}{2}$, so the other point is in the second quadrant.

Thus, the points are in quadrants $I$ and $II$, so the answer is $\boxed{\textbf{(C)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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