Question 1174470
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There is a remarkable formula to calculate the distance from a given point to a given straight line in a coordinate plane.


    Let the straight line in a coordinate plane is defined in terms of its linear equation 

         a*x + b*y + c = 0,

    where "a", "b" and "c" are real numbers, and let P = ({{{x[0]}}},{{{y[0]}}}) be the point in the coordinate plane. 

    Then the distance from the point P to the straight line is equal to

        d = {{{abs(a*x[0] + b*y[0] + c)/sqrt(a^2 + b^2)}}}.


Regarding this formula, see the lesson
    <A HREF=https://www.algebra.com/algebra/homework/Vectors/The-distance-from-a-point-to-a-straight-line-in-a-coordinate-plane.lesson>The distance from a point to a straight line in a coordinate plane</A>
in this site.


Your straight line is 2x + 3y - 6 = 0.   The point is  (4,-2)


Substitute the given data  a= 2, b= 3, c= -6,  {{{x[0]}}} = 4,  {{{y[0]}}}= -2  into the formula to get the distance under the question


    {{{abs(2*4 + 3*(-2) - 6)/sqrt(2^2 + 3^2)}}} = {{{abs(-4)/sqrt(13)}}} = {{{4/sqrt(13)}}} = {{{(4*sqrt(13))/13}}} = 1.1094  (approximately).


<U>Answer</U>.  The distance is  {{{(4*sqrt(13))/13}}} = 1.1094  (approximately).
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Solved.