Question 1079374
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<pre>
The line passing through the two given points has the equation

{{{(y-(-3))/(x-2)}}} = {{{(-3-3)/(2-5)}}},   or

{{{(y+3)/(x-2)}}} = {{{(-6)/(-3)}}},   or

{{{(y+3)/(x-2)}}} = 2,   which is the same as

y+3 = 2*(x-2),   or

2x - y -7 = 0.

Having this equation, you can check, that your points lie in this straight line.


   (From your course of analytic geometry you must know the procedure I applied above . . . )


On what to do next, see the lesson
&nbsp;&nbsp;&nbsp;&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>
in this site.


You will find there everything you need to complete the solution on your own.



       OK. The last step is this formula for the distance, taken from that lesson:

             d = {{{abs(2*12 + (-1)*7 - 7)/sqrt(2^2 + (-1)^2)}}} = {{{10/sqrt(5)}}} = {{{2*sqrt(5)}}}.
</pre>


Also, &nbsp;you have this free of charge online textbook on Geometry

&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A> 

in this site.


The referred lesson is the part if this textbook under the topic 
"<U>The distance from a point to a straight line in a coordinate plane</U>".



Please let me know whether this lesson was useful to you.  
Thank you.