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Math Help - slope and line

  1. #1
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    Post slope and line

    A rectangle ABCD, where A(3,2) and B(1,6).
    (i) Find the equation of BC.
    (ii)Given that the equation of AC is y=x-1, Find the coordinates of C.
    (iii) the perimeter of the rectangle ABCD. PLEASE HELP!!!
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  2. #2
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    Hello

    It might be helpful to start by plotting the points and the line you are given.

    Can you find the equation of a line from two points? Do you know how to find the gradient of a line perpendicular to another line?

    The point C must lie on the line which is perpendicualr to AB and passes through point B (1,6).

    Using the points A and B you can find the equation of AB and then BC, see above.

    AC is a diagonal of the rectangle and you can find the point C by finding the intersection of the lines AC and BC.

    If you plot the co-ordinates of A, B and C you might be able to see how to find the perimeter.

    Can you proceed from here. Please ask if you need more help.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by kaley015 View Post
    A rectangle ABCD, where A(3,2) and B(1,6).
    (i) Find the equation of BC.
    (ii)Given that the equation of AC is y=x-1, Find the coordinates of C.
    (iii) the perimeter of the rectangle ABCD. PLEASE HELP!!!
    (i)
    The equation of BC can be found by the point-slope form of the equation of the line.
    (y -y1) = m(x -x1)

    You have the point (x1,y1) = B(1,6)
    So find the m.
    m is the negative reciprocal of the slope of side AB.

    You should get the equation of BC as.
    (y -6) = (1/2)(x -1)
    y = (1/2)x +(11/2) ------------answer.

    ----------------------
    (ii)Given that the equation of AC is y=x-1, Find the coordinates of C.

    Get the intersection of
    y = (1/2)x +11/2 ----------(1)
    and
    y = x -1 ------------------(2)
    to find the coordinates of point C

    You should get C(13,12) ----------answer.

    ---------------------------------------------
    (iii) the perimeter of the rectangle ABCD.

    The perimeter of ABCD is twice the sum of AB +BC,
    P = 2(AB +CD)

    Distance, d, between two points is
    d = sqrt[(x2 -x1)^2 +(y2 -y1)^2]

    So for AB,
    AB = sqrt[(1 -3)^2 +(6 -2)^2] = sqrt(20) = 2sqrt(5)

    You should get perimeter of ABCD = 16sqrt(5) ------answer.
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