Question

If the area of the triangle formed by the lines $y=x$, $x+y=2$ and the line through $P(h,k)$ and parallel to $x$-axis is $4h^2$, the locus of $P$ can be

• A. $2x-y+1=0$
• B. $2x+y-1=0$
• C. $x-2y+1=0$
• D. $x+2y-1=0$

Answer 1

Answer: (A), (B)

$2x+y-1=0$

Coordinates of $A$ are $(1,1)$ which is the point of intersection of the given lines. $y=k$ is the line through $P$ parallel to $x$-axis which meets the given lines at $B$ and $C$. So coordinates of $B$ are $(k,k)$ and $C$ are $(2-k,k)$.
Area of $\Delta ABC=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ k& k& 1\\ 2-k& k& 1\end{array}\right|=4h^2$
$\Rightarrow (k-1{)}^2=4h^2\Rightarrow k-1=\pm 2h$
Locus of $P(h,k)$ is $y-1=\pm 2x$
$\Rightarrow$ $2x-y+1=0$ or $2x+y-1=0$