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finding the equation of a perpendicular line watch

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    Q: Find the equation of the straight line which passes through the point (0,3) and is perpendicular to the straight line with equation y=2x.

    My answer:

    y=2x

    y=-(1/2)x + 3

    Have I done this right? I'm not sure about the y-intercept, although it's given in the question.
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    (Original post by frostyy)
    Q: Find the equation of the straight line which passes through the point (0,3) and is perpendicular to the straight line with equation y=2x.

    My answer:

    y=2x

    y=-(1/2)x + 3

    Have I done this right? I'm not sure about the y-intercept, although it's given in the question.
    I think you are correct
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    ans is 2y=6-x which is also y=3-0.5x, yep you are correct.
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    Yeah? What about this question, because it's making me mad.

    The graphs of y = 2x^2 and y = mx - 2 intersect at the point A and B. The point B has coordinates (2, 8).

    Find the coordinates of the point A.


    The graph looks kind of like this:


    No clue where to start
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    (Original post by frostyy)
    Yeah? What about this question, because it's making me mad.

    The graphs of y = 2x^2 and y = mx - 2 intersect at the point A and B. The point B has coordinates (2, 8).

    Find the coordinates of the point A.


    The graph looks kind of like this:


    No clue where to start
    You're given co-ordinates in the question - what can you do with them?
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    (Original post by SeanFM)
    You're given co-ordinates in the question - what can you do with them?
    well I worked out from the coordinates that the gradient of line y = mx - 2 is 5, so it's y = 5x - 2.

    I'm not sure what else I can do.
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    (Original post by frostyy)
    well I worked out from the coordinates that the gradient of line y = mx - 2 is 5, so it's y = 5x - 2.

    I'm not sure what else I can do.
    That's good, that would've been your first step after knowing how to answer the question.

    Soo.. how do you answer the question?

    If two lines intersect...
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    When you've 2 equations what do you do to find the intersection point(s)?
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    (Original post by frostyy)
    well I worked out from the coordinates that the gradient of line y = mx - 2 is 5, so it's y = 5x - 2.

    I'm not sure what else I can do.
    Okay, so now you know that A is one of the points where the line y=5x-2 and the curve y=2x^2 intersect. You can use this to find the co-ordinates of A.
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    Good replies, answering me with what my question was if I'd knew how to find coordinates of an intercept I wouldn't ask.
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    Well was trying to help you out through it.. Anws you have to equate both equations to find the x values. One will be 2 (which you already know) and the other one will be the x coordinate of A.
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    (Original post by frostyy)
    Good replies, answering me with what my question was if I'd knew how to find coordinates of an intercept I wouldn't ask.
    When two lines intersect, you can find the points of intersection by making the equations of the lines equal to each other.

    Basically, find all x such that 2x^2 = 5x - 2 to find the x coordinates of the points of intersection.
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    I don't understand.


    Have I done this correctly?

    Q: Express as a single algebraic fraction:

    ( 1 / x - 2 ) - ( 2 / x + 4)

    A:

    ( 1 + 6 / x - 4 + 6) - (2 / x + 4)

    ( 7 / x + 4 ) - ( 2 / x + 4)

    ( 7 - 2 ) / ( x + 4 + x + 4)

    = 5 / ( 2x + 8 )
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    (Original post by frostyy)
    I don't understand.


    Have I done this correctly?

    Q: Express as a single algebraic fraction:

    ( 1 / x - 2 ) - ( 2 / x + 4)

    A:

    ( 1 + 6 / x - 4 + 6) - (2 / x + 4)

    ( 7 / x + 4 ) - ( 2 / x + 4)

    ( 7 - 2 ) / ( x + 4 + x + 4)

    = 5 / ( 2x + 8 )
    I don't think you have. Are you trying to simplify \displaystyle \frac {1} {x-2} - \frac {2} {x + 4} ? And can you use LaTeX please? It's easier to read.
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    (Original post by frostyy)
    I don't understand.


    Have I done this correctly?

    Q: Express as a single algebraic fraction:

    ( 1 / x - 2 ) - ( 2 / x + 4)

    A:

    ( 1 + 6 / x - 4 + 6) - (2 / x + 4)

    ( 7 / x + 4 ) - ( 2 / x + 4)

    ( 7 - 2 ) / ( x + 4 + x + 4)

    = 5 / ( 2x + 8 )

    At the point A, the y-values of the line and the curve are both the same. So it is (one of) the point(s) where 5x-2=2x^2. This is a quadratic that you can solve to get the values of x, and then sub back in to get y.
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    (Original post by frostyy)
    I don't understand.
    By the way, I just want to explain the last problem to you again. You have two lines: y=2x^2 and y=5x-2. To find the points that they meet, you just make the lines equal to each other like so: 2x^2=5x-2. Solve for x to get the x coordinates of the intersecting points.
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    (Original post by frostyy)
    I don't understand.


    Have I done this correctly?

    Q: Express as a single algebraic fraction:

    ( 1 / x - 2 ) - ( 2 / x + 4)

    A:

    ( 1 + 6 / x - 4 + 6) - (2 / x + 4)

    ( 7 / x + 4 ) - ( 2 / x + 4)

    ( 7 - 2 ) / ( x + 4 + x + 4)

    = 5 / ( 2x + 8 )
    Are there brackets around x-2 and x+4?
    Also, you can't get an equivalent fraction by adding or subtracting a number on either side of a fraction.
    So by adding 6 to the top and bottom of the fraction changes the fraction.
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    (Original post by Dingooose)
    By the way, I just want to explain the last problem to you again. You have two lines: y=2x^2 and y=5x-2. To find the points that they meet, you just make the lines equal to each other like so: 2x^2=5x-2. Solve for x to get the x coordinates of the intersecting points.
    Further to this post, I would like to add that the reason why you make the two lines equal to each other is because they will share the same x and y at the point of intersection. Therefore, using that the y's will be equal, you can equate the two equations.
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    (Original post by Dingooose)
    By the way, I just want to explain the last problem to you again. You have two lines: y=2x^2 and y=5x-2. To find the points that they meet, you just make the lines equal to each other like so: 2x^2=5x-2. Solve for x to get the x coordinates of the intersecting points.
    Solving this really got me messed up:
    2x^2=5x-2

    What I did:
    2x^2-5x+2=0
    (x+2)(2x+1)=0

    so x = -2 and x = -1/2 ?

    I tried to do it with quadratics, but I'm pretty sure this is not the correct way to do it.
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    (Original post by frostyy)
    Solving this really got me messed up:
    2x^2=5x-2

    What I did:
    2x^2-5x+2=0
    (x+2)(2x+1)=0

    so x = -2 and x = -1/2 ?

    I tried to do it with quadratics, but I'm pretty sure this is not the correct way to do it.
    You're absolutely on the right lines, but you factorised the quadratic incorrectly. Double check the signs.
 
 
 
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