+0  
 
0
115
8
avatar

The line y = (x-3)/2 intersects the circle x^2 + y^2 at A and B. Find the coordinates of the midpoint of AB.

 

This is real urgent thank you

 Jun 14, 2020
 #1
avatar+8340 
0

Sorry, but the circle equation is not complete. It should be x^2 + y^2 = (missing part). What is that missing part?

 Jun 14, 2020
 #2
avatar
0

Oops typo. it's x^2 + y^2 = 5

Guest Jun 14, 2020
 #3
avatar+338 
+4

I think https://www.khanacademy.org might help

 Jun 14, 2020
 #4
avatar
+1

I actually got the answer I just wanna double check. I have (3/5, -6/5)

 Jun 14, 2020
 #5
avatar+8340 
0

First, you substitute y = (x - 3)/2 into x^2 + y^2 = 5.

 

You will get \(x^2 + \left(\dfrac{x - 3}2\right)^2 = 5\)

 

Now upon simplification, \(5x^2 - 6x - 11 = 0\).

 

Notice that the solutions to this equation is the x-coordinates of A and x-coordinates of B respectively.

 

We conclude that the x-coordinate of the midpoint of AB is the sum of roots divided by 2, which is \(\dfrac35\).

 

By a similar approach, you can find that the y-coordinate of the midpoint is \(-\dfrac65\), which means your answer is correct.

 Jun 14, 2020
 #7
avatar
+1

ah ok. Thank you

Guest Jun 14, 2020
 #6
avatar
0

Nvm I double checked and I got the correct one. Thank you

 Jun 14, 2020
 #8
avatar+1128 
+1

well we get slope 1/2 and we find points -1,2 and 2.2,0.4 to inersect w ethen use the pythangorean theorem to find the coorinates of midpoint AB is 3/5, -6/5

 Jun 14, 2020

20 Online Users

avatar
avatar