+0  
 
+1
404
1
avatar

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+7266 
+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
 #1
avatar+7266 
+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

8 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.