Question 1149120
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<pre>

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 - y + 3 = 0.


Substitute the given data  a= 2, b= -1, c= 3,  {{{x[0]}}} = -6,  {{{y[0]}}}= 6  into the formula to get the distance under the question


    {{{abs(2*(-6) + (-1)*6 + 3)/sqrt(2^2+(-1)^2)}}} = {{{abs(-15)/sqrt(5)}}} = {{{15/sqrt(5)}}} = {{{(15*sqrt(5))/5}}} = {{{3*sqrt(5)}}} = 6.71  (approximately - rounded as requested).


<U>Answer</U>.  The distance is  {{{3*sqrt(5)}}} = 6.71  (approximately - rounded as requested).



    {{{drawing( 400, 400, -8, 8, -8, 8,
            grid(1), 
            line (-7, -11,  7, 17),
            locate (8, 2.8, ax+by+c=0),

            circle (-6,   6,   0.15),
            locate (-6.6, 6.6, P),
            locate (-3, 5.4, d),
        red(line(-6, 6, 0, 3))
)}}}

    Straight line &nbsp;{{{a*x+b*y+c=0}}},
    the point &nbsp;<B>P</B>({{{x[0]}}},{{{y[0]}}})&nbsp; and the distance {{{d}}}
    from the point &nbsp;<B>P</B>&nbsp; to the straight line
</pre>

Solved.