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Math Help - Find the coordinates...

  1. #1
    Member Mr Rayon's Avatar
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    Find the coordinates...

    Find the coordinates of the point(s) of intersection of the line y = 2x + 5 with the circle (x - 3)^2 + (y + 4)^2 = 50.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Where are you stuck ?
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  3. #3
    Member Mr Rayon's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Where are you stuck ?
    From the beginning. I don't know what to do.

    Do you think you can provide me a complete worked out solution to this?
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  4. #4
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    Well you have two equations and two unknowns. Start by plugging in y = 2x + 5 into the other equation, like this:
    (x - 3)^2 + ({\color{red}2x + 5} + 4)^2 = 50.

    Solve for x, and when you're done, plug that back into y = 2x + 5 to solve for y. (You may get two answers for x, so plug each answer in to get a different y.)


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  5. #5
    Member Mr Rayon's Avatar
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    (x - 3)^2 + (2x + 5 + 4)^2 = 50
    (x - 3)^2 + (2x + 9)^2 = 50
    x^2 - 6x + 9 + 4x^2 + 36x + 81 = 50
    5x^2 + 30x + 90 = 50
    5x^2 + 30x + 40 = 0

    Using the quadratic formula:

    x = -2 or x = -4

    y = 2 (-2) + 5
    = 1

    or

    y = 2(-4) + 5
    = -3

    Am I right so far? If so, what do I do next?
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  6. #6
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    Quote Originally Posted by Mr Rayon View Post
    x = -2 or x = -4

    y = 2 (-2) + 5
    = 1

    or

    y = 2(-4) + 5
    = -3

    Am I right so far? If so, what do I do next?
    That's it. You have the points: (-2, 1) and (-4, -3).

    BTW, you didn't have to use the quadratic formula -- the quadratic was factorable.

    \begin{aligned}<br />
5x^2 + 30x + 40 &= 0 \\<br />
5(x^2 + 6x + 8) &= 0 \\<br />
5(x + 2)(x + 4) &= 0 \\<br />
...<br />
 \end{aligned}


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