<p>A straight line passing through the point <img alt="$(0,4)$" height="18" src="https://latex.artofproblemsolving.com/5/5/9/559c5f1a789e7c0b85a47cea72d5217259ecb924.png" width="38" /> is perpendicular to the line <img alt="$x-3y-7=0$" height="16" src="https://latex.artofproblemsolving.com/2/f/8/2f8d717ffbac714f52f6a1b56b9df6b89a8a17bf.png" width="115" />. Its equation is:</p> <p><img alt="$\text{(A) } y+3x-4=0\quad \text{(B) } y+3x+4=0\quad \text{(C) } y-3x-4=0\quad \\ \text{(D) } 3y+x-12=0\quad \text{(E) } 3y-x-12=0$" height="37" src="https://latex.artofproblemsolving.com/3/0/f/30f7135705443aeb670e78ae6b00264cd3847c07.png" width="472" /></p>

Question No: 457

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

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

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Tags: Geometry

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