CEL
Last Activity: 7 Years ago
A(12,8) x1,y1
B(-2,6) x2,y2
C(6,0) x3,y3
We need to show that AB2=BC2+AC2
AB2= (x2-x1)2+(y2-y1)2
AB= sqrt(200)
similarly find the distance of BC and AC with the distance formula
BC2= (6+2)2+(0-6)2
BC= 10
AC2= (12-6)2+(8-0)2
AC= 10
so triangle ABC is a right angled triangle through pythagoras theorem.
Midpoint Q of hypotenuse AB = 12-2/2 , 8+6/2
=(5,7) x4,y4
We need to show that AQ=QB=QC
QC2=(x4-x3)2+(y4-y3)2
QC2=(5-6)2+(7-0)2
QC= sqrt(50)
AQ=QB
AQ=AB/2
=sqrt(200)/2
=2sqrt(50)/2
=sqrt(50)
therefore AQ=BQ=QC
hence proved :)