How do you find the equation of a line that is perpendicular to y=2/3x-4y=23x−4 and goes through point (6,-2)?

1 Answer
Massimiliano
Jan 29, 2015

The answer is: y=-3/2x+7y=−32x+7

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

y=mx+qy=mx+q

where

mm is the slope of the line and
qq is the y-intercept.

The formula that gives the equation of a line rr if its slope, m_rmr, and a point PP that in r∈r are known is:

y-y_P=m_r(x-x_P)y−yP=mr(x−xP)

Two lines r and srands are perpendicular if m_r=-1/m_smr=−1ms.

So the slope of the requested line is: m_r=-1/(2/3)=-3/2mr=−123=−32.

The equation of the line is:

y-(-2)=-3/2(x-6)rArry+2=-3/2x+9rArry=-3/2x+7y−(−2)=−32(x−6)⇒y+2=−32x+9⇒y=−32x+7.

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