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    1) i) The straight line p is perpendicular to the line with equation x+2y=1 and passes through the point A (a,2). Find, in terms of the constant a, an equation for the line p.

    1) ii) Given that the line p crosses the y-axis at the point B (0,3), find the value of a, and hence find the distance AB.

    For part i) I found the gradient of line p to be 2 and came up with the final answer as 2x-y=-2(1-a), is this correct? As for part ii) I found a to be -1/2 but I can't proceed any further until I know that I'm getting the right answers.

    Thanks for any help
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    (Original post by jordanwu)
    1) i) The straight line p is perpendicular to the line with equation x+2y=1 and passes through the point A (a,2). Find, in terms of the constant a, an equation for the line p.

    1) ii) Given that the line p crosses the y-axis at the point B (0,3), find the value of a, and hence find the distance AB.

    For part i) I found the gradient of line p to be 2 and came up with the final answer as 2x-y=-2(1-a), is this correct? As for part ii) I found a to be -1/2 but I can't proceed any further until I know that I'm getting the right answers.

    Thanks for any help
    Correct but a very untidy way of writing the equation

    You will not be able to check partial answers in an exam
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    (Original post by TenOfThem)
    Correct but a very untidy way of writing the equation

    You will not be able to check partial answers in an exam
    Sorry but what do you mean by untidy? I found the distance AB to be
    sqr(5/4).
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    (Original post by jordanwu)
    Sorry but what do you mean by untidy? I found the distance AB to be
    sqr(5/4).
    I mean untidy

    Why not write it as y =

    Why have that -2 on the right hand side



    Correct for AB though I would square root the 4
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    (Original post by TenOfThem)
    I mean untidy

    Why not write it as y =

    Why have that -2 on the right hand side



    Correct for AB though I would square root the 4
    So now I'm stuck on this question :/

    Two straight lines have equations x+3y=13a and 3x-y=9a+10, where a is a constant. Their point of intersection is denoted by A.
    Find, in terms of a, the coordinates of A.

    How would I start as there are 3 variables!
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    (Original post by jordanwu)
    So now I'm stuck on this question :/

    Two straight lines have equations x+3y=13a and 3x-y=9a+10, where a is a constant. Their point of intersection is denoted by A.
    Find, in terms of a, the coordinates of A.

    How would I start as there are 3 variables!
    Do as you would normally do

    The third variable is a and will be in the answers
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    (Original post by TenOfThem)
    Do as you would normally do

    The third variable is a and will be in the answers
    So I got (4a+3, 3a-1)?
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    (Original post by jordanwu)
    So I got (4a+3, 3a-1)?
    Why the

    ?

    If you put those into the equations do they work?
 
 
 
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