Question 1185927
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Reduce the equation 4x + 3y + 20 = 0 to the normal form and find the distance of the line from the origin
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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 4x + 3y + 20 = 0.


Substitute the given data  a= 4, b= 3, c= 20,  {{{x[0]}}} = 0,  {{{y[0]}}}= 0  into the formula to get the distance under the question


    {{{abs(4*0 + 3*0 + 20)/sqrt(4^2+3^2)}}} = {{{abs(20)/sqrt(25)}}} = {{{20/5}}} = 4.     <U>ANSWER</U>.


<U>Answer</U>.  The distance from the line to the origin is 4 units.
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Solved.