MathDB

3

Part of 1968 AMC 12/AHSME

Problems(1)

Equation of a Line

Source: 1968 AHSME Problem #3

8/31/2011
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:
<spanclass=′latex−bold′>(A)</span> y+3x−4=0<spanclass=′latex−bold′>(B)</span> y+3x+4=0<spanclass=′latex−bold′>(C)</span> y−3x−4=0<spanclass=′latex−bold′>(D)</span> 3y+x−12=0<spanclass=′latex−bold′>(E)</span> 3y−x−12=0<span class='latex-bold'>(A)</span>\ y+3x-4=0 \qquad <span class='latex-bold'>(B)</span>\ y+3x+4=0 \qquad <span class='latex-bold'>(C)</span>\ y-3x-4=0 \qquad\\ <span class='latex-bold'>(D)</span>\ 3y+x-12=0 \qquad <span class='latex-bold'>(E)</span>\ 3y-x-12=0
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