What is the distance between #(5, –1)# and #(3,7)#?

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Answer 1 · 2016-02-10T21:36:05

Use the distance formula: #d = sqrt((y_2-y_1)^2+(x_2-x_1)^2)#

This yields a distance of #sqrt 68# units.

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

# = sqrt((7-(-1))^2 +(3-5)^2) = sqrt(64+4) = sqrt 68#

Answer 2 · 2016-02-10T21:39:52

#"distance exactly"=2sqrt(17)" "#
#"distance approximately"~= 8.25" to 2 decimal places"#

Tony B

Now consider them as forming a triangle:
Tony B

From this you can see that Pythagoras will give us the answer for the distance between the points.

Let distance be #d# then

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

so #d = sqrt((-8)^2+(2)^2#

so#" " d= sqrt(68) = sqrt(2^2xx17)#

#=2sqrt(17)#