Question

The area of the triangle formed by the intersection of a line parallel to $x-$axis and passing through $P(h,k)$ with the lines $y=x$ and $x+y=2$ is $4h^2$. Then the locus of the point $P$ is

• A. $y=3x-2$ or $y=-3x-2$
• B. $y=2x-1$ or $y=-2x-1$
• C. $y=3x+2$ or $y=-3x+2$
• D. $y=2x+1$ or $y=-2x+1$

Answer 1

The correct option is D $y=2x+1$ or $y=-2x+1$

A line passing through $P(h,k)$ and parallel to $x-$ axis is $y=k$ ........................$\left(1\right)$
The other two given lines are
$y=x$ ........................$\left(2\right)$
and $x+y=2$ .........................$\left(3\right)$
Solving eqns $\left(1\right)$ and $\left(2\right)$
we get $A(k,k)$
Solving eqns $\left(2\right)$ and $\left(3\right)$
we get $B(1,1)$
Solving eqns $\left(3\right)$ and $\left(1\right)$
we get $C(2-k,k)$
$\therefore$ Area of $△ABC=4h^2$
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}k& k& 1\\ 1& 1& 1\\ 2-k& k& 1\end{array}\right|=4h^2$
Applying $C_2\to C_2-C_1,$ then
$\frac{1}{2}\left|\begin{array}{ccc}k& 0& 1\\ 1& 0& 1\\ 2-k& 2k-2& 1\end{array}\right|=4h^2$
$\Rightarrow {(k-1)}^2=4h^2$
$\Rightarrow k-1=\pm 2h$
$\therefore k=\pm 2h+1$
Replace $h\to x,k\to y$
$\therefore$ Locus of $(h,k)$ is $y=2x+1$ or $y=-2x+1$