How do you write an equation of the line in standard form that is Perpendicular to 3y-2x=6 and through (-1,2)?

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Answer 1 · 2018-05-19T02:06:09

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

The first step to solve this is to transform the original equation into #y=mx+b# form.

#3y-2x=6#
→Add #2x# to both sides

→#3y=2x+6#
→Divide #3# on both sides

→#y=2/3x+2#

A perpendicular line to this equation means that the slope of the new line will be the opposite of the original equation's slope.

For example if the original equation was #y=2x+3#, the slope of the perpendicular line would be #-1/2#.

So for this problem, the slope of perpendicular line will be #-3/2#.

Now that you have the slope, you have to solve for the #b#-value:

#y=-3/2x+b#
→ #2=-3/2(-1)+b#
→ #2=3/2+b#
→Subtract #3/2# to both sides
→#1/2=b#

Now that you have both the #b# and the #m#-value, you plug them into the #y=mx+b# form to get your answer.

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