Help, write an equation... ?

I asked this earlier but wrote the question wrong. Write an equation in standard form for a line that is (a) parallel (b) perpendicular to the line with an equation of y = 3x - 5 that passes through the point (8, 5).   

3 Answers

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  • 8 months ago

    (a) y = 3x + 5 - 3×8 which is y = 3x - 19

    (b) y = (-1/3)x + 5 + 8/3 which is y = -⅓x + ²³∕₃

    Attachment image
    Source(s): The line parallel to y=mx+b through a given point (x₀,y₀) is y = mx + y₀ - mx₀, while the line perpendicular to y=mx+b through a given point (x₀,y₀) is y = (-1/m)x + y₀ + x₀/m.
  • 8 months ago

    Equation of a straight line with a slope m and given point( a, b )

    y - b = m( x - a )

    y = 3x - 5

    slope: m₁ = 3

    parallel lines: have same slope ===> m₁ = m₂ 

    perpendicular lines: product of their slopes is negative one ===>  m₁m₂ = -1

    ( a )Parallel: 

    m₁ = m₂ = 3

    ( a, b ) = ( 8, 5 )

    y - 6 = 3( x - 8 )

    y - 6 = 3x - 24

    y = 3x - 18

    ——————

    Perpendicular: 

    m₁m₂ = -1

    m₂ = -⅓ 

    ( a, b ) = ( 8, 5 )

    y - 6 = -⅓( x - 8 )

    y - 6 = -⅓x - 8/3

    y = -⅓x + 10/3

    —————— 

  • Tasm
    Lv 6
    8 months ago

    y = 3x - 5

    parallel just change the last number

    y = 3x - 6

    perpendicular just reverse the sign of the x value

    y = -3x - 5

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