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Math Help - [SOLVED] College Algebra: the distance formula

  1. #1
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    [SOLVED] College Algebra: the distance formula

    Solve the problem.

    Find all the points having an x-coordinate of 9 whose distance from the point (3,-2) is 10.

    A) (9,13), (9,-7)
    B) (9, -12), (9,8)
    C) (9,6),(9,-10)
    D) (9,2), (9,-4)

    List the intercepts for the graph of the equation.

    x^2+15x+54

    A) (6,0), (9,0),(0,54)
    B) (0,6), (0,9), (54,0)
    C) (0,-6), (0,-9), (54,0)
    D) (-6,0), (-9,0), (0,54)

    This is off a test review (doesn't count only to study), so I have the answers. I just need help solving them. If someone could please help me out.
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  2. #2
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    Quote Originally Posted by Ashley189 View Post
    Solve the problem.

    Find all the points having an x-coordinate of 9 whose distance from the point (3,-2) is 10.

    A) (9,13), (9,-7)
    B) (9, -12), (9,8)
    C) (9,6),(9,-10)
    D) (9,2), (9,-4)

    List the intercepts for the graph of the equation.

    x^2+15x+54

    A) (6,0), (9,0),(0,54)
    B) (0,6), (0,9), (54,0)
    C) (0,-6), (0,-9), (54,0)
    D) (-6,0), (-9,0), (0,54)

    This is off a test review (doesn't count only to study), so I have the answers. I just need help solving them. If someone could please help me out.
    1. distance formula ...

    10 = \sqrt{(9-3)^2 + (y+2)^2}

    solve for y ... you should get two solutions.



    2. y-intercept ... evaluate y = x^2 + 15x + 54 for x = 0.

    x-intercepts ...

    x^2 + 15x + 54 = 0

    (x + 9)(x + 6) = 0

    know what to do now?
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  3. #3
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    Quote Originally Posted by skeeter View Post
    1. distance formula ...

    10 = \sqrt{(9-3)^2 + (y+2)^2}

    solve for y ... you should get two solutions.



    2. y-intercept ... evaluate y = x^2 + 15x + 54 for x = 0.

    x-intercepts ...

    x^2 + 15x + 54 = 0

    (x + 9)(x + 6) = 0

    know what to do now?
    For two I got it.

    x^2 + 15x + 54 = 0

    (x + 9)(x + 6) = 0

    x+9=0 or x+6=0
    x=-9 x=-6

    So, the answers D.

    For #1 I'm still stuck.
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  4. #4
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    10 = \sqrt{(9-3)^2 + (y+2)^2}

    start by squaring both sides ...

    100 = (9-3)^2 + (y+2)^2

    100 = 36 + (y + 2)^2

    64 = (y+2)^2

    \pm 8 = y+2

    -2 \pm 8 = y
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  5. #5
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    Quote Originally Posted by skeeter View Post
    10 = \sqrt{(9-3)^2 + (y+2)^2}

    start by squaring both sides ...

    100 = (9-3)^2 + (y+2)^2

    100 = 36 + (y + 2)^2

    I get it up to here ^

    64 = (y+2)^2

    \pm 8 = y+2

    -2 \pm 8 = y
    I don't know how to do it from that point on.
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  6. #6
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    Quote Originally Posted by Ashley189 View Post
    I don't know how to do it from that point on.
    Not sure if this is what you want to know .

    -2\pm8=y , there will be 2 equations .

    -2+8=y or -2-8=y

    There will be 2 values of y ie 6 and -10
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  7. #7
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    Now I got it. Didn't know that +- was two separate equations that's the part that got me confused.
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