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    At this moment of time I feel like strangling co-ordinative geometry
    but I dont think thats very realistic
    I am do pracitse questions and the mark scheme does not show how they got to the answer.
    So could you awesomely clever people show how they got these answers.

    The points A(1,7) B (20,7) C(15,-3) form the verticies of a triangle ABC, as shown in Figure 1. The point D (8,2) is the midpoint of AC

    The line l , which passes through D and is perpendicular to AC , interscets AB at E.
    a)find the equation for l , in the form ax+by+c=0 where a,b and c are intergers.
    b) Find the exact co-ordinates of e


    Thank you guys so much,

    positive rep for the one which is most helpful
    Thanksssssssssssss
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    a) find the gradient of the line AD. then, find gradient of normal to this. then, use this and the coordinates of D to find the line
    b) find the equation of the line AB, and solve simultaneously with the solution of a)
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    You can find the line through A and C by using the points themselves.

    You are then told that l is perpendicular (hint: what does this do to the gradient?) and that l passes through D so use these two facts to give your answer to a)

    For b), the question is merely "What is the point of intersection of l and AB?"
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    Well first find the gradient of the line AC using the points of the triangle i.e.:
    

\frac{15-1}{-3-7}=-\frac{7}{5}
    Now if the line is perpendicular to AC then the gradients of AC and l multiplied together give -1. So you just divide -1 by the gradient of AC to get:
    

\frac{5}{7}
    This is the gradient of l. Now you know it passes through D. So use the formula for finding the equation of the line with only one point.
    

y-2=\frac{5}{7}(x-8)
    Now simplify this into the form required in the question.
    For the next part work out the equation of the line of AB by working out its gradient and then using either A or B with the line formula in the C1 book(the one which only uses one point and the gradient). Solve simulataneously with the equation of l
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    you can make it easier for yourself by drawing a diagram really.

    well, AC, of which lies midpoint, D, and is perpendicular to l1.
    so find gradient of AC, which is y2-y1/x2-x1. you should get -3-7/15-1. so -10/14.
    as l1 is perpendicular to D. get the reciprocal of gradient, so 14/10
    then put it into the equation y-y1=m(x-x1) where x1 and y1 are (8,2) as the line goes through it.

    y-2=14/10(x-2)
    y-2=1.4x-11.2

    in the form of ax+by+c=0
    1.4x-y-9.2=0


    and for b)

    you know that E must have y-coordinate of 7 as both A and B have y-coordinate of 7.

    so just sub y=7 into
    1.4x-y-9.2=0
    so, 1.4x-7-9.2=0
    1.4x=16.2
    x=11.571....

    so the coordinates of E are
    (11.57,7)

    hope it helps
 
 
 
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