Given equation are y = x , y = 2 x − 1 , 3 y − 4 x = 1 y=x,y=2x-1,3y-4x=1 y = x , y = 2 x − 1 , 3 y − 4 x = 1
Let y = x . . . ( 1 ) y=x...(1) y = x ... ( 1 )
y = 2 x − 1 … ( 2 ) y=2x-1…(2) y = 2 x − 1 … ( 2 )
3 y − 4 x = 1 … ( 3 ) 3y-4x=1…(3) 3 y − 4 x = 1 … ( 3 )
Solving (1) and (2), we get:
( x , y ) = ( 1 , 1 ) (x,y)=(1,1) ( x , y ) = ( 1 , 1 )
Solving (2) and (3), we get:
3 ( 2 x − 1 ) − 4 x = 1 3(2x-1)-4x=1 3 ( 2 x − 1 ) − 4 x = 1
⇒ 2 x = 4 ⇒2x=4 ⇒ 2 x = 4
⇒ x = 2 ⇒x=2 ⇒ x = 2
∴ y = 3 ∴y=3 ∴ y = 3
( x , y ) = ( 2 , 3 ) (x,y)=(2,3) ( x , y ) = ( 2 , 3 )
Solving (1) and (3), we get:
( x , y ) = ( − 1 , − 1 ) (x,y)=(-1,-1) ( x , y ) = ( − 1 , − 1 )
So, let coordinates of points A, B, C of △ A B C \triangle ABC △ A BC be ( 1 , 1 ) , ( 2 , 3 ) , ( − 1 , − 1 ) (1,1),(2,3),(-1,-1) ( 1 , 1 ) , ( 2 , 3 ) , ( − 1 , − 1 ) respectively.
We know that area of triangle with vertices ( x 1 , y 1 ) , ( x 2 , y 2 ) , ( x 3 , y 3 ) (x_1,y_1),(x_2,y_2),(x_3,y_3) ( x 1 , y 1 ) , ( x 2 , y 2 ) , ( x 3 , y 3 ) is given as:
△ = ∣ 1 2 ∣ x 1 y 1 1 x 2 y 2 1 x 3 y 3 1 ∣ ∣ \triangle =|\frac{1}{2}
\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix}| △ = ∣ 2 1 ∣ ∣ x 1 x 2 x 3 y 1 y 2 y 3 1 1 1 ∣ ∣ ∣
∴ \therefore ∴ Area of △ A B C = ∣ 1 2 ∣ 1 1 1 2 3 1 − 1 − 1 1 ∣ ∣ \triangle ABC=|\frac{1}{2}
\begin{vmatrix}
1 & 1 & 1 \\
2 & 3 & 1 \\
-1 & -1 & 1
\end{vmatrix}| △ A BC = ∣ 2 1 ∣ ∣ 1 2 − 1 1 3 − 1 1 1 1 ∣ ∣ ∣
= 1 2 [ 1 ( 3 + 1 ) − 1 ( 2 + 1 ) + 1 ( − 2 + 3 ) ] =\frac{1}{2} [1(3+1)-1(2+1)+1(-2+3)] = 2 1 [ 1 ( 3 + 1 ) − 1 ( 2 + 1 ) + 1 ( − 2 + 3 )]
= 1 2 [ 4 − 2 + 1 ] =\frac{1}{2} [4-2+1] = 2 1 [ 4 − 2 + 1 ]
= 3 2 =\frac{3}{2} = 2 3 square units.
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