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Find the equation of a line, passes thro...

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

Text Solution

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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`
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Knowledge Check

  • Equation of a line passing through (-1,2,-3) and perpendicular to the plane 2x+3y+z+5=0 is

    A
    `(x-1)/(-1)=(y+2)/1=(z-3)/(-1)`
    B
    `(x+1)/(-1)=(y-2)/1=(z+3)/1`
    C
    `(x+1)/2=(y-2)/3=(z+3)/1`
    D
    none of these

  • Equation of a line passing through (1,-2) and perpendicular to the line 3x-5y+7=0 is

    A
    5x+3y+1=0
    B
    3x+5y+1=0
    C
    5x-3y-1=0
    D
    3x-5y+1=0

  • Equatin of the line passing through (-1, 1) and perpendicular to the line 2x +3y +4 =0 is

    A
    `2(y-1)=3(x+1)`
    B
    `3 (y-1) =-2 (x+1)`
    C
    `y-1 =2 (x+1)`
    D
    `3 (y-1) = x+1`

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    Equation of line passing through the point (1,2) and perpendicular to the line y=3x-1 is

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