Question 675162
to show that a triangle with vertices (5,-3) (2,-6) and (1,1) is a right angled triangle, first find the length of each side which is actually the distance between each given point  

(5,-3) and (2,-6) 

*[invoke Distance_Formula_for_Coordinate_Plane 5, -3, 2, -6]


(5,-3)and (1,1)

*[invoke Distance_Formula_for_Coordinate_Plane 5, -3, 1,1]


(2,-6) and (1,1)

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



{{{a=4.24}}}

{{{b=5.67}}}


{{{c=7.07}}}

if {{{c^2=a^2+b^2}}}, then a triangle is a right angled triangle


{{{c^2=a^2+b^2}}}


{{{7.07^2=4.24^2+5.66^2}}}


{{{49.99=17.98+32}}}

{{{49.99=49.98}}}

{{{50=50}}}.............so, a triangle is a right angled triangle

{{{drawing( 600, 600, -10, 10, -10, 10, 
         grid(1),
         (triangle( 5, -3, 2, -6, 1, 1 )), 
  (locate(5,-3,A(5,-3))), locate(2,-6,B(2,-6)), locate( 1,1,C(1,1))) ) )
)}}}



 the area; 

{{{a=4.24}}}.......height

{{{b=5.67}}}.......base

{{{A=a*b/2=(4.24*5.67)/2=24.0408/2=12.0204=12}}}