Question 981367
You meant (6,1), (2,-3), and (4,-5), which indeed are the vertices of a right triangle.
{{{drawing(300,300,-2,8,-7,3,
grid(1),
triangle(6,1,2,-3,4,-5),
locate(6.1,1.5,A(6,1)),
locate(4,-5,B(4.1,-5)),
locate(0.1,-2.5,C(2,-3)),
locate(3.1,-3.5,a),
locate(3.5,-1.5,b),
locate(4.5,-2,c)
)}}}


ONE WAY:
Show that the line connecting (6,1) and (2,-3) has a slope of {{{1}}}.
Show that the line connecting (4,-5) and (2,-3) has a slope of {{{-1}}}.
Since the product of the slopes is {{{(1)(-1)=-1}}} , the lines are perpendicular,
so the triangle has a right angle at (2,-3).
The slope the line connecting {{{A(6,1)}}} and {{{C(2,-3)}}} is
{{{(1-(-3))/(6-2) =(1+3)/4=4/4=1}}}
The slope the line connecting {{{B(4,-5)}}} and {{{C(2,-3)}}} is
{{{(-5-(-3))/(4-2) =(-5+3)/2=-2/2=-1}}}


ANOTHER WAY:
Show that the lengths of the sides satisfy the Pythagorean theorem:
{{{c^2=a^2+b^2}}}
{{{c^2=(6-4)^2+(1-(-5))^2=2^2+6^2=4+36=40}}}
{{{a^2=(2-4)^2+(-3-(-5))^2=(-2)^2+2^2=4+4=8}}}
{{{b^2=(6-2)^2+(1-(-3))^2=4^2+4^2=16+16=32}}}