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

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(B) Given,the pair of lines is $x^2-y^2+2y-1=0$. $x^2 = y^2-2y+1$ $\Rightarrow x^2 = (y-1)^2$ $\Rightarrow x^2 - (y-1)^2 = 0$ $\Rightarrow (x+y-1)(x-y

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(B) Given,the pair of lines is $x^2-y^2+2y-1=0$. $x^2 = y^2-2y+1$ $\Rightarrow x^2 = (y-1)^2$ $\Rightarrow x^2 - (y-1)^2 = 0$ $\Rightarrow (x+y-1)(x-y+1) = 0$ So,the lines are $l_1: x+y-1=0$ and $l_2: x-y+1=0$. The angle bisectors of these lines are found by $\frac{x+y-1}{\sqrt{1^2+1^2}} = \pm \frac{x-y+1}{\sqrt{1^2+(-1)^2}}$. $\Rightarrow x+y-1 = \pm(x-y+1)$. Case $1$: $x+y-1 = x-y+1$ $\Rightarrow 2y = 2$ $\Rightarrow y=1$. Case $2$: $x+y-1 = -(x-y+1)$ $\Rightarrow x+y-1 = -x+y-1$ $\Rightarrow 2x = 0$ $\Rightarrow x=0$. The angle bisectors are the lines $x=0$ ($Y$-axis) and $y=1$. The triangle is formed by the lines $x+y=4$,$x=0$,and $y=1$. The vertices of the triangle are: $1$. Intersection of $x=0$ and $y=1$ is $(0, 1)$. $2$. Intersection of $x=0$ and $x+y=4$ is $(0, 4)$. $3$. Intersection of $y=1$ and $x+y=4$ is $(3, 1)$. The base of the triangle along $x=0$ is $|4-1| = 3$. The height of the triangle from $x=0$ to $x=3$ is $3$. Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 3 	imes 3 = 4.5$ sq. units.

The area of the triangle formed by the line $x+y=4$ and the angle bisectors of the pair of lines $x^2-y^2+2y-1=0$ is ......... sq. units.

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