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the points A(2,3) B(4,-1) C(-1,2) are the vertices of a triangle. find the length and perpendicular from A to BC and hence the area of ABC

  • maths -

    a. Find the equation of the line BC, given two points (B and c)

    Then, find the perpendicular (given the negative inverse slope from a), and the point A.

    Area? Area=lengthBC*lengthperpendicular*1/2

    Just finding the area, there are other ways easier.

    Graph it, use Pick's Theorem http://en.wikipedia.org/wiki/Pick%27s_theorem
    or
    http://www.mathopenref.com/coordtrianglearea.html

    or to check your answer ONLY: http://www.gottfriedville.net/mathtools/triarea.html

  • maths -

    consider BC as your base:
    BC = √(5^2 + (-3)^2) = √34
    slope of BC = -3/5

    so the slope of the perpendicular = 5/3
    equation of line from A to BC , using A as the point
    y-3 = (5/3)(x-2)
    3y - 9 = 5x - 10
    5x - 3y = 1 ---- #1

    equation of BC , using C
    y-2 = -(3/5)(x+1)
    5y - 10 = -3x -3
    3x + 5y = 7 ----#2

    5 times #1 --- 25x - 15y = 5
    3 times #2 --- 9x + 15y = 21
    add them
    34x = 26
    x = 26/34 = 13/17 .... ughh
    in #2
    3(13/17) + 5y = 7
    ..
    y = 16/17 , yuk, the point on BC = (13/17 , 16/17)
    so length of altitude
    = √(2-13/17)^ + (3-16/17)^2)
    = √(441/289 + 1125/289) = √(1666/289)
    = 7√34/17

    so the area = (1/2) base x heigh
    = (1/2) * √34 * (7/17)√34 = 7 , yeahhhh


    There are of course much easier ways to find the area of a triangle if you are given the three points.

    The simplest way is to list the 3 points in a column repeating the first one you listed.

    2 3
    4 -1
    -1 2
    2 3

    area = (1/2) | ( sum of downproducts - sum of upproducts)|
    = (1/2)|(-2+ 8 - 3 -(12 + 1 + 4)|
    = (1/2)| -14 |
    = 7

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