+0  
 
+1
54
1
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Alice is on the line x=4, and she is  √40 away from (2,3). What is the sum of the possible y-coordinates of the point Alice is at?

 

Please help! I think the answer is 0 but I'm not sure.

Guest Aug 19, 2018
edited by Guest  Aug 19, 2018

Best Answer 

 #1
avatar+7266 
+1

Alice is one the line  x = 4  ,  so the  x-coordinate of whatever point Alice is on is  4 .

Let's call the point that Alice is located  (4, y)  .

 

By the distance forumula...

 

the distance between  (4, y)  and  (2, 3)   =   \(\sqrt{(4-2)^2+(y-3)^2}\)

 

The problem tells us that the distance between  (4, y)  and  (2, 3)  is  √40 ,  so...

 

\(\sqrt{(4-2)^2+(y-3)^2}=\sqrt{40}\)

                                                          Now let's solve this equation for  y .  First square both sides.

\((4-2)^2+(y-3)^2=40\)

                                                          Simplify  \((4-2)^2\)  to  \(4\) .

\(4+(y-3)^2=40\)

                                                          Subtract  4  from both sides of the equation.

\((y-3)^2=36\)

                                                          Take the ± square root of both sides.

\(y-3=\pm\sqrt{36}\)

 

\(y-3=\pm6\)

                                                          Add  3  to both sides.

\(y=3\pm6\)

 

\(\begin{array}{ccc} y=3+6&\qquad\text{or}\qquad &y=3-6\\~\\ y=9&\text{or}&y=-3 \end{array}\)

 

The possible y-coordinates of the point Alice at are  9  and  -3 .

 

The sum of the possible y-coordinates of the point Alice is at   =   9 + -3   =   6

 

To help check the answer, we can see on this graph that  (4, 9)  and  (4, -3) , the two possible points Alice is at, are the same distance away from  (2, 3) ...over 2 units and up or down  6  units.

hectictar  Aug 19, 2018
 #1
avatar+7266 
+1
Best Answer

Alice is one the line  x = 4  ,  so the  x-coordinate of whatever point Alice is on is  4 .

Let's call the point that Alice is located  (4, y)  .

 

By the distance forumula...

 

the distance between  (4, y)  and  (2, 3)   =   \(\sqrt{(4-2)^2+(y-3)^2}\)

 

The problem tells us that the distance between  (4, y)  and  (2, 3)  is  √40 ,  so...

 

\(\sqrt{(4-2)^2+(y-3)^2}=\sqrt{40}\)

                                                          Now let's solve this equation for  y .  First square both sides.

\((4-2)^2+(y-3)^2=40\)

                                                          Simplify  \((4-2)^2\)  to  \(4\) .

\(4+(y-3)^2=40\)

                                                          Subtract  4  from both sides of the equation.

\((y-3)^2=36\)

                                                          Take the ± square root of both sides.

\(y-3=\pm\sqrt{36}\)

 

\(y-3=\pm6\)

                                                          Add  3  to both sides.

\(y=3\pm6\)

 

\(\begin{array}{ccc} y=3+6&\qquad\text{or}\qquad &y=3-6\\~\\ y=9&\text{or}&y=-3 \end{array}\)

 

The possible y-coordinates of the point Alice at are  9  and  -3 .

 

The sum of the possible y-coordinates of the point Alice is at   =   9 + -3   =   6

 

To help check the answer, we can see on this graph that  (4, 9)  and  (4, -3) , the two possible points Alice is at, are the same distance away from  (2, 3) ...over 2 units and up or down  6  units.

hectictar  Aug 19, 2018

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