How do I write the equation of a line that is perpendicular to $y = 3 x + 4$ $y=3x+4$ and goes through the point (3,5)?

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How do I write the equation of a line that is perpendicular to $y = 3 x + 4$ $y=3x+4$ and goes through the point (3,5)?

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Answer 1 · 2015-11-21T05:29:07

$y = - \frac{1}{3} x + 6$ $y=-1/3x+6$

Explanation:

Two perpendicular lines have opposite reciprocal slopes.

The slope in $y = 3 x + 4$ $y=3x+4$ is $3$ $3$ , so the line perpendicular to it will have a slope of $- \frac{1}{3}$ $-1/3$ .

We can write this in the form $y = - \frac{1}{3} x + b$ $y=-1/3x+b$ and then plug in $(3, 5)$ $(3,5)$ for $x$ $x$ and $y$ $y$ to find the equation of the line perpendicular to the line $y = 3 x + 4$ $y=3x+4$ that goes through the point $(3, 5)$ $(3,5)$ .

$5 = - \frac{1}{3} (3) + b$ $5=-1/3(3)+b$
$5 = - 1 + b$ $5=-1+b$
$6 = b$ $6=b$

Therefore, the equation of our new line is: $y = - \frac{1}{3} x + 6$ $y=-1/3x+6$ .

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How do I write the equation of a line that is perpendicular to $y = 3 x + 4$ $y=3x+4$ and goes through the point (3,5)?

1 Answer

Explanation:

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How do I write the equation of a line that is perpendicular to $y = 3 x + 4$ $y=3x+4$ and goes through the point (3,5)?