Question 1100406
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There is a beatiful formula to find the distance from the given point to the given straight line in a coordinate plane.


This formula is:


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


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


    where &nbsp;<B>a</B>, &nbsp;<B>b</B> and &nbsp;<B>c</B> &nbsp;are real numbers, &nbsp;and let &nbsp;<B>P</B> = <B>P</B>({{{x[0]}}},{{{y[0]}}})&nbsp; is the point in the coordinate plane 
    with the coordinates &nbsp;{{{x[0]}}}, &nbsp;{{{y[0]}}}. &nbsp;Then the distance from the point &nbsp;<B>P</B>&nbsp; to the straight line is equal to


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


See the lesson &nbsp;<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>&nbsp; in this site.



So, in your case  &nbsp;a= -1, &nbsp;b= 2, &nbsp;c= -14   &nbsp;&nbsp;(after moving &nbsp;"c" &nbsp;from the righ side to the left), &nbsp;{{{x[0]}}}= -14,  &nbsp;{{{y[0]}}}= 5.


Substituting to the formula gives you the distance


    d = {{{abs((-1)*(-14) + 2*5 -14)/sqrt((-1)^2+2^2)}}} = {{{abs(14+10-14)/sqrt(1 + 4)}}} = {{{10/sqrt(5)}}} = {{{(10*sqrt(5))/5)}}} = {{{2*sqrt(5)}}} = 4.5 (approximately; rounded to the nearest tenth).



Solved.



Again, &nbsp;for the proof of the formula and other details look into the above referred lesson.