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