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

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(B) The given pair of straight lines is $x^2-y^2+2y=1$.This can be rewritten as $x^2-(y^2-2y+1)=0$,which simplifies to $x^2-(y-1)^2=0$.Factoring this,

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$2$

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(B) The given pair of straight lines is $x^2-y^2+2y=1$.This can be rewritten as $x^2-(y^2-2y+1)=0$,which simplifies to $x^2-(y-1)^2=0$.Factoring this,we get $(x-(y-1))(x+(y-1))=0$,so the lines are $x-y+1=0$ and $x+y-1=0$.The equations of the angle bisectors 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 $1$: $x-y+1 = x+y-1$ $\Rightarrow 2y=2$ $\Rightarrow y=1$.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 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 triangle is a right-angled triangle with base of length $2$ (from $x=0$ to $x=2$ along $y=1$) and height of length $2$ (from $y=1$ to $y=3$ along $x=0$).Area of the triangle = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 2 = 2$ sq. units.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$K$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

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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