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Find the sum of all values of $a$ such that the point $(a,7)$ is $3\sqrt{5}$ from the point $(2,1)$.

Guest Nov 19, 2017

Best Answer 

 #1
avatar+7052 
+1

From the Pythagorean theorem...

 

[ distance between (a, 7)  and  (2, 1) ]2   =   [a - 2]2  +  [7 - 1]2

 

And they tell us that the distance equals  3√5  , so

 

[  3√5 ]2   =   [a - 2]2  +  [7 - 1]2

 

45   =   [a - 2]2  +  36

 

9   =   [a - 2]2

 

±√9  =   a - 2

 

± 3   =   a - 2

 

± 3  +  2   =   a

 

a  =  3 + 2  =  5          or          a = -3 + 2  =  -1

 

and

 

5  +  -1   =   4

hectictar  Nov 19, 2017
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1+0 Answers

 #1
avatar+7052 
+1
Best Answer

From the Pythagorean theorem...

 

[ distance between (a, 7)  and  (2, 1) ]2   =   [a - 2]2  +  [7 - 1]2

 

And they tell us that the distance equals  3√5  , so

 

[  3√5 ]2   =   [a - 2]2  +  [7 - 1]2

 

45   =   [a - 2]2  +  36

 

9   =   [a - 2]2

 

±√9  =   a - 2

 

± 3   =   a - 2

 

± 3  +  2   =   a

 

a  =  3 + 2  =  5          or          a = -3 + 2  =  -1

 

and

 

5  +  -1   =   4

hectictar  Nov 19, 2017

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