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HELP PLZ!!!! DUE TODAY

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

Guest Jun 14, 2020
#3
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Jun 14, 2020
#4
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I actually got the answer I just wanna double check. I have (3/5, -6/5)

Jun 14, 2020
#5
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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
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ah ok. Thank you

Guest Jun 14, 2020
#6
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Nvm I double checked and I got the correct one. Thank you

Jun 14, 2020
#8
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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