Question

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

Answer 1

☆Answer☆

if lines are perpendicular

then product of slopes =-1

let slope of required line be m

○ Slope of 2x+3y-6=0 is -2/3

so m×(-2/3)=-1

m=3/2

so required line is 3x-2y+c=0

now substitute x=-1 and y=1

3(-1)-2(1)+c=0

c=5

Required line = 3x-2y+5=0

hope it helps

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