1968 AHSME Problems/Problem 3

Problem

A straight line passing through the point $(0,4)$ is perpendicular to the line $x-3y-7=0$. Its equation is:

$\text{(A) } y+3x-4=0\quad \text{(B) } y+3x+4=0\quad \text{(C) } y-3x-4=0\quad \\ \text{(D) } 3y+x-12=0\quad \text{(E) } 3y-x-12=0$

Solution

The original line can be adjusted to standard form, in which it is $x-3y=7$. Because this line is in standard form, its slope is $\frac{-1}{-3}=\frac{1}{3}$. Thus, the slope of the perpendicular line is the negative reciprocal of this number, $-3$. Because we know that this new line passes through the point $(0,4)$, we can describe this line using point-slope form, in which it is $y-4=-3(x-0)$, or $y+3x-4$, which is answer choice $\fbox{A}$.

See also

1968 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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