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vegas14

  • 4 years ago

Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, −1).

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  1. 14yamaka
    • 4 years ago
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    y=1/3x-3

  2. mark_o.
    • 4 years ago
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    3x + y = 7 y=-3x+7 here slope m = -3 the line perpendicular has slope m=-1/m=-1/(-3)=1/3 now at point(6,-1)=(x1,y1) from y-y1=m(x-x1) we get y-(-1)=(1/3)(x-6) y+1 = (1/3)x -2 y= (1/3)x -2-1=(1/3)x-3

  3. amistre64
    • 4 years ago
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    SWAP COEFFS AND NEGATE ONE Of em ..... bad pinky!! 3x+y = 7 x-3y = 7 , that sevens from something other than what we want so delete in and find a new constant x-3y = c the given point has to fit ...

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