+0

Find all values of A

+1
43
1

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

#1
+5573
+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
+5573
+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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