Question 1137655
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You want to find the distance from the point (5,0) to the given straight line  0.5x - y + 5 = 0.


There is a remarkable formula which ideally suits for this need.


    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.


Substitute the given data  a= 0.5, b= -1, c= 5,  {{{x[0]}}} = 5,  {{{y[0]}}}= 0  into the formula to get the distance under the question


    {{{abs(0.5*5 + (-1)*0 + 5)/sqrt(0.5^2+(-1)^2)}}} = {{{7.5/1.118}}} = 6.708.


<U>Answer</U>.  The distance under the question is 6.708 units.
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Solved.