The area bounded by the angle bisectors of th | vedclass

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(A) The given equation is ${x^2} - ({y^2} - 2y + 1) = 0$,which simplifies to ${x^2} - {(y - 1)^2} = 0$.This represents the pair of lines $(x - (y - 1)

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$2$

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(A) The given equation is ${x^2} - ({y^2} - 2y + 1) = 0$,which simplifies to ${x^2} - {(y - 1)^2} = 0$.This represents the pair of lines $(x - (y - 1))(x + (y - 1)) = 0$,or $(x - y + 1)(x + y - 1) = 0$.The lines are $x - y + 1 = 0$ and $x + y - 1 = 0$.The angle bisectors of these lines are given by $\frac{x - y + 1}{\sqrt{1^2 + (-1)^2}} = \pm \frac{x + y - 1}{\sqrt{1^2 + 1^2}}$.This simplifies to $x - y + 1 = \pm (x + y - 1)$.Case $1$: $x - y + 1 = x + y - 1 \implies 2y = 2 \implies y = 1$.Case $2$: $x - y + 1 = -(x + y - 1) \implies x - y + 1 = -x - y + 1 \implies 2x = 0 \implies x = 0$.Thus,the angle bisectors are the lines $x = 0$ and $y = 1$.The third line is $x + y = 3$.The vertices of the triangle formed by $x = 0$,$y = 1$,and $x + y = 3$ are found by intersection:Intersection of $x = 0$ and $y = 1$ is $(0, 1)$.Intersection of $x = 0$ and $x + y = 3$ is $(0, 3)$.Intersection of $y = 1$ and $x + y = 3$ is $(2, 1)$.The triangle is a right-angled triangle with vertices $(0, 1)$,$(0, 3)$,and $(2, 1)$.The base length is $2 - 0 = 2$ and the height is $3 - 1 = 2$.Area = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 2 = 2$.

The area bounded by the angle bisectors of the lines ${x^2} - {y^2} + 2y = 1$ and the line $x + y = 3$ is:

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