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Math Help - Coordinate geometry

  1. #1
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    Lightbulb Coordinate geometry

    The points A(6,7), B(0,1) and C(9,4) are vertices of a triangle.Find the length AB and the area of the triangle ABC and hence obtain the perpendicular distance from C to the line AB.
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  2. #2
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    Surely you can find the distance AB, can't you? The best way of figuring the area depends on what you covered about the topic. One way is to use cross product: the area is one half of the length of the cross product of the vectors AB and AC.

    Please describe in more detail what you know and what your difficulty is.
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  3. #3
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    how do we find the area using only the length AB?
    how do we use the area to find the perpendicular distance?
    Isn't the perpendicular distance the height of the triangle?
    why is it 'hence' then?
    Note: I haven't studied vectors as yet!

    Thanku 4 ur help!
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  4. #4
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    how do we find the area using only the length AB?
    Of course, you need more than AB to find the area. I said that you can find AB because the problem explicitly asks for it and because I was surprised that you could not find the distance between two points with known coordinates. (See the paragraph about the Euclidean plane in this Wikipedia article.)

    The forum rules ask for the poster to show some effort in solving the problem, so finding AB seems to be a perfect way to do this.

    how do we use the area to find the perpendicular distance?
    The area is half the product of a side and the height towards this side.

    You still have to say what ways of finding the area you studied. The easiest way in this case is this formula, but you need to shift the triangle so that A is at the origin (0, 0).
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  5. #5
    Super Member bigwave's Avatar
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    Cool use distance forumula

    If (x_1, y_1) and (x_2,y_2) are any two points and if d is the distance between them, then

    d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}

    thus if A(6,7) and B(0,1) then

    \overline{AB} = \sqrt{(6-0)^2 + (7-1)^2}<br />
\Rightarrow \sqrt{36 + 36}<br />
\Rightarrow \sqrt{72}<br />
\Rightarrow 6\sqrt{2}<br />

    also, the slope intercept form of \overline{AB} is y=x+1 where m=1 thus the perpendicular to \overline{AB} is the negative recipical of 1 which is -1

    Since \overline{AC} has m= -1 you can use that for h

    thus \overline{AC} = \sqrt{(9-6)^2+(4-7)^2} \Rightarrow 3\sqrt{2}

    then \overline{AB} is b then use

    A = \frac{1}{2}bh<br />
\Rightarrow<br />
\frac{1}{2} (3\sqrt{2}) (6\sqrt{2}) = 36


    draw this on graph paper then you can what is happening
    try to take some steps then ask where you are stuck
    Last edited by bigwave; October 19th 2010 at 12:05 PM. Reason: latex
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