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 $\fbox{A}$ .

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1968 AHSME Problems/Problem 3

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