Question 391969
<pre><font size = 4 color = "indigo" face = "batangche"><b>

There are two ways:

First draw the two points and draw a line segment from one to the
other:

{{{drawing(400,400,-7, 7,-10,4, graph(400,400,-7,7,-10,4),

green(line(4,-1,-2,-5)), locate(-4,-5,"(-2,-5)"), locate(4,-.5,"(4,-1)")
)}}}

Draw a right triangle whose hypotenuse is the green line:

{{{drawing(400,400,-7, 7,-10,4, graph(400,400,-7,7,-10,4),
red(line(-2,-5,4,-5),line(4,-5,4,-1)),
green(line(4,-1,-2,-5)), locate(-4,-5,"(-2,-5)"), locate(4,-.5,"(4,-1)")
)}}}

Count the units in the lower red leg of the right triangle and you'll find it
is 6 units long.

Count the units in the right red leg of the right triangle and you'll find it 
is 4 units long.

Use the Pythagorean theorem and find the length of the green hypotenuse:

{{{c^2=a^2+b^2}}}
{{{c^2=6^2+4^2}}}
{{{c^2=36+16}}}
{{{c^2=52}}}
{{{c=sqrt(52)}}}
{{{c=sqrt(4*13)}}}
{{{c=2sqrt(13)}}}

(4,-1) and (-2,-5).

Another way is to use the formula, which amounts to the same thing:

{{{d=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

{{{d=sqrt(((-2)-(4))^2+((-5)-(-1))^2)}}}

{{{d=sqrt((-2-4)^2+(-5+1)^2)}}}

{{{d=sqrt((-6)^2+(-4)^2)}}}

{{{d=sqrt(36+16)}}}

{{{d=sqrt(52)}}}

{{{d=sqrt(4*13)}}}

{{{d=2sqrt(13)}}}

Edwin</pre></font></b>