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| Area of a triangle |
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| If the coordinates of the vertices of the DABC are A(x1,y1), B(x2,y2) and C(x3,y3), then the area of triangle is given by |
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| Draw AL, BM and CN perpendicular to x-axis. |
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| LM = x2-x1 |
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| MN = x3-x2 |
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| LN = x3-x1 |
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| Area of D ABC = area of trap {ALMB + BMNC - ALNC} |
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| This can be expressed in the form of a determinant |
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| Another form which is convenient to use for the area of triangles but which is very much useful when areas of polygons of more than three sides are to be determined is |
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| Here the first row x1, y1 is repeated in the last row. |
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| Note: |
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| If three points are collinear, then the area of the triangle is zero. |
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| This is the condition for collinearity of three points. |
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| Example: |
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| Show that the points (a,b+c), (b,c+a) and (c,a+b) are collinear. |
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| Suggested answer: |
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| If the area of the figure formed by the three coordinates is zero, then the three points are collinear. |
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| \The three given points are collinear. |
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