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Find the equation of the line perpendicular to 2x+3y-6=0 and passing through (-1,1)

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Answer:
y-1=(3)/(2)(x+1)

Explanation:


2x+3y-6=0\\\\3y=-2x+6\\\\y=-(2)/(3)x+2

So, the slope of the given line is -2/3, meaning the slope of the line we want to find is 3/2.

Substituting into point-slope form, we get the equation of the line is


y-1=(3)/(2)(x+1)

User Sachin Mandhare
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