Question 985219
A=({{{2}}},{{{-6}}}); B=({{{0}}},{{{-6}}}); C=({{{2}}},{{{1}}})

 Plot each point and from the right triangle {{{ABC}}}. 

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(2,-6,.12), locate(2,-6,A(2,-6)),
circle(0,-6,.12), locate(0,-6,B(0,-6)),
circle(2,1,.12), locate(2,1,C(2,1)),
line(2,-6,0,-6),line(2,-6,2,1),line(2,1,0,-6),

 graph( 600, 600, -10, 10, -10, 10, 0)) }}}

to verify that the triangle is a right triangle, find the length of each side and prove that
{{{(BC)^2=(BA)^2+(AC)^2}}}

find {{{BC}}}

*[invoke Distance_Formula_for_Coordinate_Plane 0, -6, 2, 1]

find {{{AC}}}

*[invoke Distance_Formula_for_Coordinate_Plane 2, -6, 2, 1]

find {{{BA}}}

*[invoke Distance_Formula_for_Coordinate_Plane 2, -6, 0, -6]


so, {{{BC=7.2801098828052=7.28}}}, {{{AC=7}}}, and {{{BA=2}}}

{{{(BC)^2=(BA)^2+(AC)^2}}}

{{{(7.2801098828052)^2=2^2+7^2}}}

{{{52.9984=4+49}}}

{{{53=53}}}

now find its area:

{{{A=(1/2)(BA)(AC)}}}

{{{A=(1/2)2*7}}}

{{{A=(1/cross(2))cross(2)*7}}}

{{{A=7}}}