How do you find the equation of a line that is perpendicular to $y=2/3x-4$ and goes through point (6,-2)?

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Answer 1 · 2015-01-29T20:51:53

The answer is: $y=-3/2x+7$

The equation of a line, written in explicit form is:

$y=mx+q$

where

$m$ is the slope of the line and
$q$ is the y-intercept.

The formula that gives the equation of a line $r$ if its slope, $m_r$ , and a point $P$ that $in r$ are known is:

$y-y_P=m_r(x-x_P)$

Two lines $r and s$ are perpendicular if $m_r=-1/m_s$ .

So the slope of the requested line is: $m_r=-1/(2/3)=-3/2$ .

The equation of the line is:

$y-(-2)=-3/2(x-6)rArry+2=-3/2x+9rArry=-3/2x+7$ .

How do you find the equation of a line that is perpendicular to y=2/3x-4y=2/3x-4 and goes through point (6,-2)?