Find the equation of a line, passes through `(-1,2)` and perependicular to the line `2x+3y=1`.

Explanation

Given straight line is
`2x+3y=1`
Its slope
`m_(1)=-("Coefficient of" x)/("Coefficient of" y)=-(2)/(3)`
Slope of perpendicular line `m_(2)=-(1)/(m_(1))`
`=-(1)/((2)/(3))=(3)/(2)`
Now, equation of a line passes through the point `(-1,2)` and whose slope is `(3)/(2)` is
`y-2=(3)/(2)(x+1)`
`implies 2y-4=3x+3`
`implies 3x-2y+7=0`