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Co-Ordinate Geometry.. watch

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    (Original post by apo1324)
    ok...
    y=3x-1 (x=1)
    y=3(1)-1 = 3 - 1 = 2 (1,2)

    y=3x-1 (y=2)
    2=3x-1
    2+1=3x
    3/3=x
    x=1 (1,2)



    x+3y-7=0 (x=1)
    1+3y-7=0
    3y=-1+7
    3y=+6
    y=+6/3
    y=2 (1,2)

    x+3y-7=0 (y=2)
    x+3(2)-7=0
    x+6-7=0
    x-1=0
    x=1 (1,2)

    Are they correct??
    Yeah.

    You can also sub both straight in to each equation, and the sides will equal each other.
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    what do u mean, if you don't mind me asking... is it an easier way
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    (Original post by apo1324)
    what do u mean, if you don't mind me asking... is it an easier way
    Yeah, less working.

    Just sub in both:
    y = 3x - 1
    2 = 3(1) - 1
    2 = 3 - 1
    2 = 2

    Therefore the point lies on y = 3x - 1 as it satisfies the equation. It also satisfies the other line, and as they intersect at only one point (point D, where x = 1 and y = 2) we've proved it.

    For the last bit used the midpoint formula (http://www.purplemath.com/modules/xyplane/midpt07.gif) and go from there.
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    ok thanks... i rep you... bec u were like th only one who helped!!
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    (Original post by apo1324)
    ok thanks... i rep you... bec u were like th only one who helped!!
    No worries, post back if you have any problems, and I'll help tomorrow.
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    Just about to post image of full question with working and notes for you. It'll be up in 2 mins.
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    Ok... thank you very much.. it will be a little bit more clearer!!
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    Here you go:


    Hope that helps, dont take my maths as gospel mind. I did this a year ago so I had to brush up a little bit.

    When you get the hang of it you will love it, its easy compared to the rest of it!

    I did do it neat, then somehow I did 3*3=6 hah!

    So ive done it again for you, not as neat, but ive added in red lines to split each part of the question then the blue bits divide the question into seperate parts.

    Let me know how you get on, reply to me if you want more help.
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    Thank you very, very much!!! I really appreciate it!!
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    No bother Once you've got them rules learned its easy! If you get a chance do another question just to make sure you got the hang of it. I find i'll follow the answers but wont take it in sometimes.
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    its just how they word the questions which confuses me...
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    ill rep you later.. bec ive already done one today..
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    Sometimes it helps to draw a sketch, even if its got no axis. Just to get an idea. There is only so many questions that they can ask you, and they will be worded similar.

    You will need to know and be comfortable with working out:
    Gradient m1 (y2-y1/x2-x1)
    Perpendicular gradient m2=(-1/m1)
    Prove that the gradient is perpendicular (m1 * m2 MUST = -1)
    Equations of a line (y-y_1=m(x-x_1))
    Line intersection (Simultanious Equations)
    Perpendicular Bisector this sneaks up sometimes, its just a big word to put you off. Simply work out the midpoint of the line they are talking about say AB to get point C. Then work out the perpendicular gradient of AB to get C gradient. Then use the equation of a straight line.

    Thats probly about it for co-ordinate geom.

    If I've missed anything feel free to add it on.

    And thanks for the rep, no worries.
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    btw im not very good with fractions but how did u work out:-

    -1/(-1/3) this is what i struggle with the most in maths... fractions..
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    Also do you use simultaneous equations when finding a point that intersects 2 lines?
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    I use my calculator. I wont lie. I have a CASIO fx-991ES which has natural display, its a few quid more. But its GREAT. You type in what you see and it will work it out for you.

    But, say you have
    -1/(2/3) you re-arrange it so it is -1/2 * 3 which is the same as -1*3/2 = -3/2
    You can use a simple method, flip and multiply by -1.
    3/2 flip 2/3 multiply -2/3.
    Another example
    -5/6 flip 6/5 multiply 6/5
    Remember a negative multiplied by a negative is a positive (shown in above example).

    To find a point where the two lines intersect you use simultaneous yes. Sometimes they ask for a point that lies on both lines. This is just them re arranging the question, its the same thing ( where the two points intersect) as we only deal with linear equations (AFAIK) there will only be one point of intersection.
 
 
 
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