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    • Thread Starter
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    Hi there,

    Firstly, thanks in advance for your help!.

    1) Find the equation of the line joining the points (2,-3) and (5,12)

    Is the answer y=5x+12?

    2) Find the co-ordinates of the point of intersection of the lines with equations
    3x+y=1 and y=7x-2

    I'm not sure how to start here.

    3) Find the equation of the circle with centre (3,4) which passes through the point (8,-8)
    Is the answer (x-3)^2+(y-4)^2=169

    4) Show that the line y=3x-7 is a tangent to the curve y=2x^2-5x+1
    Find the co-ordinates of the point of contact.
    I've done
    3x-7=2x^2-5x+1
    3x=2x^2-5x+8
    y=2x^2-8x+8

    Is this the correct method?

    -------

    Again, thanks a lot. I do have more questions but I figured it'd be too much in one go.

    • Community Assistant
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    To see if your equations are right, you just need to substitute the points back in.

    Let's start at question 1:

    y = 5x + 12 and point (2, -3) - does this work?
    • Thread Starter
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    (Original post by Mr M)
    To see if your equations are right, you just need to substitute the points back in.

    Let's start at question 1:

    y = 5x + 12 and point (2, -3) - does this work?
    -_- Fail by me. Didn't work out C.

    5x-13?
    • Community Assistant
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    (Original post by tehforum)
    -_- Fail by me. Didn't work out C.

    5x-13?
    That one is right now. Second one is solving a pair of simultaneous equations.
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    For number 1 :
    Find the gradient first.

    Then do Y-Y1 = M(gradient)x-x1

    For number 2 :
    3x+y=1 and y=7x-2

    Rearrange the first to get it in the form y=, then make them equal each other / move to one side / factorise to get x . Then plug in to get Y.

    Havent done 3 yet .
    • Thread Starter
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    (Original post by Mr M)
    That one is right now. Second one is solving a pair of simultaneous equations.
    Second one.

    3x+y=1
    y=7x-2

    Sub 2 into 1

    3x+7x-2=1
    10x-2=1
    10x=3
    x=0.3

    Sub then x=0.3 into 1
    3(0.3)+y=1
    0.9+y=1
    y=0.1
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    (Original post by tehforum)
    Second one.

    3x+y=1
    y=7x-2

    Sub 2 into 1

    3x+7x-2=1
    10x-2=1
    10x=3
    x=0.3

    Sub then x=0.3 into 1
    3(0.3)+y=1
    0.9+y=1
    y=0.1
    Yes, third looks correct to me too.

    You started 4 correctly but then did something weird. Form a quadratic in x and work out the discriminant - what will that tell you that is relevant?
    • Thread Starter
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    3x-7=2x^2-5x+1

    This is correct? I know you've given me instructions, but I think I may need a further budge.

    Re: discriminant is it needed? b^2-4ac > 0 = 2 real roots.
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    (Original post by tehforum)
    3x-7=2x^2-5x+1

    This is correct? I know you've given me instructions, but I think I may need a further budge.

    Re: discriminant is it needed? b^2-4ac > 0 = 2 real roots.
    You need to make your quadratic equal to 0 don't you?

    The further nudge is you don't want to find that the discriminant has 2 real roots. What do you know about tangents to curves?
    • Thread Starter
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    (Original post by Mr M)
    You need to make your quadratic equal to 0 don't you?

    The further nudge is you don't want to find that the discriminant has 2 real roots. What do you know about tangents to curves?
    3x-7=2x^2-5x+1
    3x=2x^2-5x+8
    y=2x^2-8x+8
    y=x^2-4x+4
    y= (x-2)^2

    So x= 2.

    When sub that into 3x-7
    6-7 = -1

    x =2, y=-1

    (2,-1)
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    (Original post by tehforum)
    3x-7=2x^2-5x+1
    3x=2x^2-5x+8
    y=2x^2-8x+8
    y=x^2-4x+4
    y= (x-2)^2

    So x= 2.

    When sub that into 3x-7
    6-7 = -1

    x =2, y=-1

    (2,-1)
    Why have you reintroduced y in the first section?

    Why are you substituting? Were you asked to find a pair of coordinates?

    Visualise what is going on.

    You have a positive parabola so it is U shaped.

    You have a straight line with a positive gradient like this /

    If it is a tangent there should be one point of intersection.

    Trying to solve a pair of simultaneous equations and showing there is only one solution will show it is a tangent. Alternatively, you can form a quadratic and set it equal to zero (not y). If you find the discriminant you should get one real root to show there is only one answer.

    Edit: Whoops yes you were! Your coordinates are OK. I still don't like the unnecessary y and you need to write a sentence to explain why you know you have found a tangent.
 
 
 
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