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प्रश्न
Find the area of the triangle formed by the line x + y − 6 = 0, x − 3y − 2 = 0 and 5x − 3y + 2 = 0.
उत्तर
x + y − 6 = 0 ... (1)
x − 3y − 2 = 0 ... (2)
5x − 3y + 2 = 0 ... (3)
In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.
Solving (1) and (2):
x = 5, y = 1
Thus, AB and BC intersect at B (5, 1).
Solving (1) and (3):
x = 2, y = 4
Thus, AB and CA intersect at A (2, 4).
Similarly, solving (2) and (3):
x = −1, y = −1
Thus, BC and CA intersect at C (−1, −1).
∴ Area of triangle ABC = \[\frac{1}{2}\begin{vmatrix}5 & 1 & 1 \\ 2 & 4 & 1 \\ - 1 & - 1 & 1\end{vmatrix} = 12\]
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