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
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.
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