Question 849870
<pre>
They did it the long way.  

The formula for the distance from a point to a line is

{{{d}}}{{{""=""}}}{{{abs(Ax[1]+By[1]+C)/sqrt(A^2+B^2)}}}

where the equation of the line is Ax+By+C=0 and the point is (x<sub>1</sub>,y<sub>1</sub>)

The point is (x<sub>1</sub>,y<sub>1</sub>) = (6,-3)
The line's equation is y=2x-5. we must get that in the form
Ax+By+C=0

      y=2x-5.  We subtract the entire right side from both sides
 
-2x+y+5=0

A=-2, B=1, C=5, (x<sub>1</sub>,y<sub>1</sub>) = (6,-3)

{{{d}}}{{{""=""}}}{{{abs(Ax[1]+By[1]+C)/sqrt(A^2+B^2)}}}

{{{d}}}{{{""=""}}}{{{abs(-2(6)+1(-3)+(5))/sqrt((-2)^2+(1)^2)}}}

{{{d}}}{{{""=""}}}{{{abs(-12-3+5)/sqrt(4+1)}}}

{{{d}}}{{{""=""}}}{{{abs(-10)/sqrt(5)}}}

{{{d}}}{{{""=""}}}{{{10/sqrt(5)}}}

We can rationalize the denominator by multiplying by {{{sqrt(5)/sqrt(5)}}}.

{{{d}}}{{{""=""}}}{{{10/sqrt(5)}}}{{{""*""}}}{{{sqrt(5)/sqrt(5)}}}{{{""=""}}}{{{10sqrt(5)/5}}}{{{""=""}}}{{{2sqrt(5)}}}.

That's about 4.47, so let's draw the graph as a check:

{{{drawing(400,400,-3,8,-8,3,grid(1), line(13,21,-11,-27),

circle(6,-3,0.15),circle(6,-3,0.13),circle(6,-3,0.11),circle(6,-3,0.09),circle(6,-3,0.07),circle(6,-3,0.05),circle(6,-3,0.03),circle(6,-3,0.01),

green(line(6,-3,2,-1)) )}}}

That green line looks like it's about 4.47 units in length.

Edwin</pre>