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Need help to solve As levels Maths P1 Question Watch

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    Its a 10 mark question (attached) of GCE As levels Pure Maths 1 Chapter Coordinate Geometry.
    Would appreciate any help
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    What have you tried so far?
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    (Original post by ahsanshah199)
    Its a 10 mark question (attached) of GCE As levels Pure Maths 1 Chapter Coordinate Geometry.
    Would appreciate any help
    I did some rough calculations and this is what i did
    Too bad handwriting I know
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    (Original post by ahsanshah199)
    I did some rough calculations and this is what i did
    Too bad handwriting I know
    Find the equation of the line that passes through points A and B.
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    (Original post by ahsanshah199)
    I did some rough calculations and this is what i did
    Too bad handwriting I know
    There are 4 steps:

    1. Finding co-ordinates of A and B
    2. Constructing the line which is a perpendicular bisector of AB
    3. Finding the intersections, C and D, between this perpendicular bisector and the curve
    4. Using Pythagoras to calculate the distance between the C and D

    Unfortunately, you have failed to find the co-ordinates of A and B correctly.

    x-y+2=0 \Rightarrow y=x+2

    \Rightarrow 2x^2-(x+2)^2+2x+1=0

    \Rightarrow 2x^2-(x^2+4x+4)+2x+1=0

    \Rightarrow 2x^2-x^2+2x-4x+1-4=x^2-2x-3=0

    Can you see where you went wrong? Can you carry on from there?
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    (Original post by B_9710)
    Find the equation of the line that passes through points A and B.
    Thats where im stuck at
    I m making some mistake here
    Can you solve the rest for me
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    (Original post by RDKGames)
    There are 4 steps:

    1. Finding co-ordinates of A and B
    2. Constructing the line which is a perpendicular bisector of AB
    3. Finding the intersections, C and D, between this perpendicular bisector and the curve
    4. Using Pythagoras to calculate the distance between the C and D

    Unfortunately, you have failed to find the co-ordinates of A and B correctly.

    x-y+2=0 \Rightarrow y=x+2

    \Rightarrow 2x^2-(x+2)^2+y2x+1=0

    \Rightarrow 2x^2-(x^2+4x+4)+2x+1=0

    \Rightarrow 2x^2-x^2+2x-4x+1-4=x^2-2x-3=0

    Can you see where you went wrong? Can you carry on from there?
    Brother you might be doing wrong..look putting value of y, from line one that is the straight line that intersect, into equation of curve line is wrong way
    You get it
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    (Original post by ahsanshah199)
    Thats where im stuck at
    I m making some mistake here
    Can you solve the rest for me
    Actually you don't need to do that. You need to find the midpoint of AB and find a line that passes through the midpoint and is perpendicular to the line AB. I assume you can find gradients of lines and know the conditions for lines to be perpendicular.
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    (Original post by ahsanshah199)
    Brother you might be doing wrong..look putting value of y, from line one that is the straight line that intersect, into equation of curve line is wrong way
    You get it
    What??? Are you saying substituting the equation of the line into the curve is a 'wrong' way???

    I'd hate to break it to you but you're doing the exact same thing with your working out, you're just finding what y^2 is first before substituting it into the curve's equation.

    Also, having looked over your working out, you said y^2=x^2+4 which is incorrect because
    y^2=(x+2)^2 \not= x^2+4
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    (Original post by RDKGames)
    What??? Are you saying substituting the equation of the line into the curve is a 'wrong' way???

    I'd hate to break it to you but you're doing the exact same thing with your working out, you're just finding what y^2 is first.

    Also, having looked over your working out, you said y^2=x^2+4 which is incorrect because
    y^2=(x+2)^2 \not= x^2+4
    I got it
    Gonna try it myself again
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    (Original post by ahsanshah199)
    I got it
    Gonna try it myself again
    I'm just...i don't know
    I find cords for A and B
    Solved their gradient and their midpoint
    Got gradient of Line CD
    Got equation of line CD
    I substitute equation of curve to equation of line CD
    But still im doing it the wrong way
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    (Original post by ahsanshah199)
    I'm just...i don't know
    I find cords for A and B
    Solved their gradient and their midpoint
    Got gradient of Line CD
    Got equation of line CD
    I substitute equation of curve to equation of line CD
    But still im doing it the wrong way
    Show your working. You have the right approach.
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    Can someone help me with a solution
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    (Original post by ahsanshah199)
    Can someone help me with a solution
    Show your working and I'll tell you exactly where you're going wrong.
 
 
 
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