The area (in square units) of the quadrilater | vedclass

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(A) The equations of the lines are $L_1: x+y-1=0$ and $L_2: x-y+1=0$. Solving these equations,we find the point of intersection $R(0,1)$. Let $P$ be t

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$\frac{1}{2}$

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(A) The equations of the lines are $L_1: x+y-1=0$ and $L_2: x-y+1=0$. Solving these equations,we find the point of intersection $R(0,1)$. Let $P$ be the point $(1,1)$. The length of the perpendicular from $P(1,1)$ to $L_1$ is $PS = \frac{|1+1-1|}{\sqrt{1^2+1^2}} = \frac{1}{\sqrt{2}}$. The length of the perpendicular from $P(1,1)$ to $L_2$ is $PQ = \frac{|1-1+1|}{\sqrt{1^2+(-1)^2}} = \frac{1}{\sqrt{2}}$. Since $PS = PQ = \frac{1}{\sqrt{2}}$,and the triangles $	riangle PSR$ and $	riangle PQR$ are right-angled at $S$ and $Q$ respectively,they are congruent. The area of the quadrilateral $PQRS$ is the sum of the areas of $	riangle PSR$ and $	riangle PQR$. Area $= 2 	imes 	ext{Area}(	riangle PQR) = 2 	imes (\frac{1}{2} 	imes PQ 	imes QR)$. Alternatively,since $PQRS$ is composed of two congruent right triangles with legs $PQ$ and $PS$,the area is $PQ 	imes PS = \frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}} = \frac{1}{2}$ square units.

The area (in square units) of the quadrilateral formed by the point of intersection of the lines $x+y-1=0$ and $x-y+1=0$,the point $P(1,1)$,and the feet of the perpendiculars from this point onto the lines is:

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