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

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Equation of Altitude } \mathrm{AP}: \mathrm{x}=1 \ & 	ext { Equation of Altitude } \mathrm{BP}: \mathrm{y}-0=1(\mathrm{x}+2) \ & \Rightarrow \mathrm{x}=1 \& \ & \mathrm{x}-\mathrm{y}+2=0 \ & \mathrm{P}(1,3) \ & 	ext { Area of } 	riangle \mathrm{PBC}=\frac{1}{2} 	imes 4 	imes 3=6\end{aligned}$ ^

Let ABC be the triangle such that the equations of lines $A B$ and $A C$ be $3 y-x=2$ and $x+y=2$,…

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