Question 1179955
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You can directly differentiate the equation


    2x*dx + 2y*dy -6*dx + 8*dy = 0

    (2x-6)dx + (2y+8)dy = 0

    (2x-6)dx = -(2y+8)dy

    {{{(dy)/(dx)}}} = {{{-(2x-6)/(2y+8)}}}


Now substitute the given values (coordinates) into the formula to get the slope


    {{{(dy)/(dx)}}} = {{{-(2*8-6)/(2*8+8)}}} = {{{-10/24}}} = -{{{5/12}}}.


So, now you know the point (8,8) and the slope {{{-5/12}}}.


Hence, an equation of the tangent line is


    y - 8 = {{{(-5/12)(x-8)}}}.      <U>ANSWER</U>


And you can transform it equivalently to any other form you wish.
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Solved.