The area (in sq. units) of the triangle forme | vedclass

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(B) The given pair of straight lines is $x^2-y^2+2y-1=0$.This can be written as $x^2-(y-1)^2=0$,which factors as $(x-y+1)(x+y-1)=0$.Thus,the two lines

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(B) The given pair of straight lines is $x^2-y^2+2y-1=0$.This can be written as $x^2-(y-1)^2=0$,which factors as $(x-y+1)(x+y-1)=0$.Thus,the two lines are $L_1: x-y+1=0$ and $L_2: x+y-1=0$.The angular bisectors of these lines are given by $\frac{x-y+1}{\sqrt{1^2+(-1)^2}} = \pm \frac{x+y-1}{\sqrt{1^2+1^2}}$,which simplifies to $x-y+1 = \pm(x+y-1)$.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 triangle is formed by the line $x+y=3$ and the lines $x=0$ and $y=1$.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 vertices are $(0,1), (0,3),$ and $(2,1)$.The base of the triangle along the $y$-axis is $|3-1|=2$ units.The height of the triangle from the vertex $(2,1)$ to the $y$-axis is $|2-0|=2$ units.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 2 = 2$ sq. units.

The area (in sq. units) of the triangle formed by the straight line $x+y=3$ and the angular bisectors of the pair of straight lines $x^2-y^2+2y=1$ is

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