What is the distance between #(3,5)# and #(6,2)#?

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Answer 1 · 2018-05-13T03:10:10

I tried this:

Here you can use for the distance #d# the following expression (derived from Pythagoras Theorem):

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

using the coordinates of your points:

#d=sqrt((6-3)^2+(2-5)^2)=sqrt(9+9)=sqrt(18)=4.2# units

Answer 2 · 2018-05-13T03:19:20

#d = 4.24#

First, we start with the distance formula

#d = sqrt((X_2 - X_1)^2 + (Y_2 - Y_1)^2#

Coordinates are always in #(X,Y)#

So in #(3,5)#, we'll make our #3# the #X_2#
So the #5# is the #Y_2#

This means that in #(6,2)#, the #6# is the #X_1#
And the #2# is the #Y_1#

Now we plug our #X# and #Y# into the equation

#d = sqrt((3 - 6)^2 + (5 - 2)^2#

#d = sqrt(( -3)^2 + ( 3)^2#

#d = sqrt(9 + 9)#

#d = sqrt18# #~~# #4.24#