Given equation are y=x,y=2x−1,3y−4x=1
Let y=x...(1)
y=2x−1…(2)
3y−4x=1…(3)
Solving (1) and (2), we get:
(x,y)=(1,1)
Solving (2) and (3), we get:
3(2x−1)−4x=1
⇒2x=4
⇒x=2
∴y=3
(x,y)=(2,3)
Solving (1) and (3), we get:
(x,y)=(−1,−1)
So, let coordinates of points A, B, C of △ABC be (1,1),(2,3),(−1,−1) respectively.
We know that area of triangle with vertices (x1,y1),(x2,y2),(x3,y3) is given as:
△=∣21∣∣x1x2x3y1y2y3111∣∣∣
∴ Area of △ABC=∣21∣∣12−113−1111∣∣∣
=21[1(3+1)−1(2+1)+1(−2+3)]
=21[4−2+1]
=23 square units.
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