SOLUTION: find the equation of line passing through A(-2,3) and perpendicular to the line 2x-3y+6=0

Question 1120869: find the equation of line passing through A(-2,3) and perpendicular to the line 2x-3y+6=0
Found 3 solutions by Boreal, greenestamps, josgarithmetic:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
Rewrite equation of line as 3y=2x+6 or y=(2/3)x+2
perpendicular line has negative reciprocal slope or -3/2
point-slope formula y-y1=m(x-x1) where m slope and (x1, y1) point
y-3=(-3/2)(x+2)
y-3=-(3/2)x-3
y=-3/2 x

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Here is a useful shortcut for solving this kind of problem:

(1) every line parallel to the line with equation Ax+By=C (or Ax+By+C=0) will have an equation of the form Ax+By=D (or Ax+By+D=0), where D is some constant.
(2) every line perpendicular to the line with equation Ax+By=C (or Ax+By+C=0) will have an equation of the form Bx-Ay=D (or Bx-Ay+D=0), where D is some constant.

In this example, the equation of the given line is 2x-3y+6 = 0. The equation of any line perpendicular to the given line will have an equation of the form 3x+2y=D.

To find the value of D, simply plug in the coordinates of the given point:

An equation of the line we are looking for is 3x+2y=0.
Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
A line parallel to one with equation will have slope . A line perpendicular will have slope of .

Your example question requires slope .

Starting with the point-slope equation form, the equation you want is .
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