How do you write an equation of a line passing through (-3, -2), perpendicular to #2x - 3y = 3#?

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Answer 1 · 2016-08-10T19:49:49

#y = -3/2x -6 1/2#

We need the slope of the line we have been given before we can find the slope of the new line.

Change #2x-3y=3# into the form #y = mx + c#

#2x -3 = 3y" "rArr" " 3y =2x-3#
#y = 2/3x -1. " "m_1 = 2/3#.

The slope perpendicular to this is #m_2 = -3/2#
We have the slope , #(-3/2)#and a point, #(-3,-2)#
substitute the values into

#y-y_1 = m(x-x_1)#

#y-(-2) = -3/2(x-(-3))#

#y+2 = -3/2(x+3)#

#y= -3/2x-9/2-2#

#y = -3/2x -6 1/2#