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AHSME 1968 Problem 3

    A straight line passing through the point (0,4)(0,4) is perpendicular to the line x−3y−7=0x-3y-7=0. Its equation is: (A) y+3x−4=0(B) y+3x+4=0(C) y−3x−4=0(D) 3y+x−12=0(E) 3y−x−12=0\text{(A) } y+3x-4=0 \text{(B) } y+3x+4=0 \text{(C) } y-3x-4=0 \\ \text{(D) } 3y+x-12=0 \text{(E) } 3y-x-12=0

    Dataset text NuminaMath-1.5. Quoted with source.

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