What is the distance between #(2,-14)# and #(-1,21)#?

1 Answer
Jan 19, 2016

Answer:

Distance between #=sqrt(1234)~~35.128# to 3 decimal places

Explanation:

This is treated like a triangle where the line between the points is the hypotenuse.

Tony B

The distance we are after is that of AC

Given :
#(x_1,y_1)-> (2,-14)#
#(x_2,y_2)->(-1,21)#

So by Pythagoras

#( AC)^2 = (x_2-x_1)^2+(y_2-y_1)^2#

#(AC)^2=(color(white)(.)(-1) -2)^2 + (21-(-14)color(white)(.))^2#

#(AC)^2=(-3)^2+(35)^2#

#AC=sqrt(1234)~~35.128# to 3 decimal places