How do you find the possible values for a if the points (6,a), (5,0) has a distance of #sqrt17#?

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Answer 1 · 2017-04-03T12:07:57

#a=-4# or #a=4#

The distance between two points #(x_1,y_1)# and #(x_2,y_2)# is

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

Hence, distance between two points #(6,a)# and #(5,0)# is

#sqrt((5-6)^2+(0-a)^2)# and this should be #sqrt17#

Therefore #sqrt((5-6)^2+(0-a)^2)=sqrt17#

or #(5-6)^2+(0-a)^2=17#

i.e. #1+a^2=17#

or #a^2-16=0#

or #(a+4)(a-4)=0#

i.e. #a=-4# or #a=4#