Home Competitions AHSME 1968 Problem 3
A straight line passing through the point ( 0 , 4 ) (0,4) ( 0 , 4 ) is perpendicular to the line x − 3 y − 7 = 0 x-3y-7=0 x − 3 y − 7 = 0 . Its equation is:
(A) y + 3 x − 4 = 0 (B) y + 3 x + 4 = 0 (C) y − 3 x − 4 = 0 (D) 3 y + x − 12 = 0 (E) 3 y − 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 (A) y + 3 x − 4 = 0 (B) y + 3 x + 4 = 0 (C) y − 3 x − 4 = 0 (D) 3 y + x − 12 = 0 (E) 3 y − x − 12 = 0
Dataset text NuminaMath-1.5 . Quoted with source.
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