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(D) $1$. Find the coordinates of vertex $A$ by solving the equations $3y-x=2$ and $x+y=2$. Adding them gives $4y=4$,so $y=1$. Substituting $y=1$ into

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$6$

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(D) $1$. Find the coordinates of vertex $A$ by solving the equations $3y-x=2$ and $x+y=2$. Adding them gives $4y=4$,so $y=1$. Substituting $y=1$ into $x+y=2$ gives $x=1$. Thus,$A = (1, 1)$.$2$. Find the coordinates of $B$ and $C$. Since $B$ and $C$ lie on the $x$-axis $(y=0)$,substitute $y=0$ into the line equations. For $AB$: $3(0)-x=2 \Rightarrow x=-2$. So $B = (-2, 0)$. For $AC$: $x+0=2 \Rightarrow x=2$. So $C = (2, 0)$.$3$. The altitude from $A$ to $BC$ is the line passing through $(1, 1)$ perpendicular to the $x$-axis,which is $x=1$.$4$. The altitude from $B$ to $AC$ is perpendicular to $AC$. The slope of $AC$ is $-1$,so the slope of the altitude is $1$. The equation is $y-0 = 1(x-(-2)) \Rightarrow y=x+2$.$5$. The orthocentre $P$ is the intersection of $x=1$ and $y=x+2$. Substituting $x=1$ into $y=x+2$ gives $y=3$. Thus,$P = (1, 3)$.$6$. The area of $	riangle PBC$ with base $BC$ on the $x$-axis is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. The base $BC$ length is $|2 - (-2)| = 4$. The height is the $y$-coordinate of $P$,which is $3$. Area $= \frac{1}{2} 	imes 4 	imes 3 = 6$.

Let $ABC$ be a triangle such that the equations of lines $AB$ and $AC$ are $3y-x=2$ and $x+y=2$,respectively,and the points $B$ and $C$ lie on the $x$-axis. If $P$ is the orthocentre of the triangle $ABC$,then the area of the triangle $PBC$ is equal to

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