Question

The area of the triangle formed by the lines, y = x, x + y = 2 and the line through p(h, k) and parallel to the x-axis is $4h^2$, then the point P lies on

• A. 2x - y + 1 = 0
• B. 2x - y + 3 = 0
• C. x - y + 1 = 0
• D. x + 2y - 1 = 0

Answer 1

The correct option is A 2x - y + 1 = 0

Given lines are y = x . . . (1)
x + y = 2 . . . (2)
and y = k . . . (3)
solving (1) & (2) A = (1, 1)
solving (2) & (3) B = (2 - k, k)
solving (1) & (3) C = (k, k)
given area of triangle = $4h^2$
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 2-k& k& 1\\ k& k& 1\end{array}\right|=4k^2$
$\Rightarrow (k-1{)}^2=4h^2\Rightarrow k-1=\pm 2h$
$\Rightarrow$ (h,k)lies on 2x-y+1=0 (or) 2x+y-1=0