How do you find the distance between #(-sqrt2,-sqrt2)#, #(sqrt2, 6sqrt2)#?

1 Answer
Jun 13, 2017

Answer:

#abs((-sqrt2,-sqrt2):(sqrt2,6sqrt2))=color(red)(sqrt(106)#

Explanation:

The easiest way to do this is to observe that the given points are equivalent to #(-1,-1)# and #(1,6)# scaled by a factor of #sqrt(2)#

The distance between the given points will be the same as the distance between #(-1,-1)# and #(1,6)# multiplied by #sqrt(2)#

The horizontal distance between #(-1,-1)# and #(1,6)# is #1-(-1)=2#

The vertical distance between #(-1,-1)# and #(1,6)# is #6-(-1)=7#

The diagonal distance between #(-1,-1)# and #(1,6)# is given by the Pythagorean Theorem as
#color(white)("XXX")d=sqrt(2^2+7^2)=sqrt(4+49)=sqrt(53)#

and the corresponding diagonal distance between #(-sqrt(2),-sqrt(2))# and #(sqrt(2),6sqrt(2))# is
#color(white)("XXX")sqrt(2)d=sqrt(2)*sqrt(53)=sqrt(106)#