The area of the triangle formed by the line x | vedclass

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(C) The given equation is $x^2 - y^2 + 2y = 1$,which can be written as $x^2 - (y^2 - 2y + 1) = 0$,or $x^2 - (y - 1)^2 = 0$.This represents the pair of

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$2$

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(C) The given equation is $x^2 - y^2 + 2y = 1$,which can be written as $x^2 - (y^2 - 2y + 1) = 0$,or $x^2 - (y - 1)^2 = 0$.This represents the pair of lines $(x - y + 1)(x + y - 1) = 0$.The angle bisectors of these lines are given by the lines $x = 0$ and $y = 1$.The triangle is formed by the lines $x = 0$,$y = 1$,and $x + y = 3$.The vertices of the triangle are the intersection points of these lines:$1$. Intersection of $x = 0$ and $y = 1$ is $(0, 1)$.$2$. Intersection of $x = 0$ and $x + y = 3$ is $(0, 3)$.$3$. Intersection of $y = 1$ and $x + y = 3$ is $(2, 1)$.The base of the triangle along $y = 1$ is $|2 - 0| = 2$.The height of the triangle along $x = 0$ is $|3 - 1| = 2$.Therefore,the area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 2 = 2$.

The area of the triangle formed by the line $x + y = 3$ and the angle bisectors of the pair of lines $x^2 - y^2 + 2y = 1$ is:

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