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find the distance between  A(2,4) and the line  2x-3y + 5 = 0

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find the distance between  A(2,4) and the line  2x-3y + 5 = 0

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A(2,4) 

d: 2x-3y + 5 =0

The distance between a point and line formula is:

d = l (ax1 + by1 + c l/ sqrt(a^2 + b^2)

   = l (2*2 + 4*-3 + 5 l / sqrt(4 + 9)

  = l -3l / sqrt(13)

= 3/sqrt13

Then the distance is 3/sqrt13 units.

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