Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

what is the equation of a line in slope-intercept form that is perpendicular to the line y=-3x+2 and passes through (3,-1)?

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  • 2 months ago
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    Perpendicular lines have negative-reciprocal slopes.  The slope of the first line is -3 so the slope of its perpendicular is 1/3.

    We have the slope, an x, and a y so we can solve for the unknown intercept:

    y = mx + b

    -1 = (1/3)(3) + b

    -1 = 1 + b

    -2 = b

    Your equation is:

    y = (1/3)x - 2

  • 2 months ago

    Start with point-slope form. The slope of a perpendicular line is the negative reciprocal of the slope of the original line.

    So,y+1=1/3(x-3)

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

    y=(1/3)x-2

  • 2 months ago

    y = - 3x + 2

    solving the slope that is perpendicular to y = - 3x + 2

    m = - 3

    m2 = - 1/(-3) = 1/3

    solving for y - intercept using slope intercept form formula.

    y = mx + b

    - 1 = 1/3(3) + b

    - 1= 1 + b

    b = - 1 - 1

    b = - 2

    The equation of the line that is perpendicular to y = - 3x + 2 is 

    y = 1/3x - 2  Answer// 

  • ?
    Lv 7
    2 months ago

    Any line perpendicular to y = -3x+2 must have slope m = ¹⁄₃

    The equation of a line with slope m = ¹⁄₃ and passing through (3,-1) is

                   y - (-1)  = ¹⁄₃ (x - 3)

                   y = ¹⁄₃ x - 2   .......................ANS

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  • 2 months ago

    The typical equation of a line is: y = mx + y₀ → where m: slope and where y₀: y-intercept

    y = - 3x + 2 ← this is the line (ℓ₁) → the slope is (- 3)

    Two lines are perpendicular if the product of their slope is (- 1).

    As the slope of the line (ℓ₁) is (- 3), the slope of the perpendicular line (ℓ₂) is (1/3).

    The equation of the perpendicular line (ℓ₂) becomes: y = (1/3).x + y₀

    The line (ℓ₂) passes through (3 ; - 1), so the coordinates of this point must verify the equation of the line (ℓ₂).

    y = (1/3).x + y₀

    y₀ = y - (1/3).x → you substitute x and y by the coordinates of the point (3 ; - 1)

    y₀ = - 1 - [(1/3) * 3]

    y₀ = - 1 - 1

    y₀ = - 2

    The equation of the line (ℓ₂) is: y = (1/3).x + 2

  • ?
    Lv 7
    2 months ago

    y = x/3 + c

    -1 = 3/3 + c

    y = x/3 - 2

  • 2 months ago

    The perpendicular gradient is the negative reciprocal of '-3' , which is '1/3' 

    Displacing the point against (x,y) 

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

    y + 1 = x/3 - 3/3 

    y + 1 = x/3 - 1 

    y = x/3 + 2  or y = (1/3)x  + 2 

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