Question

The area of the region bounded by locus of $P$ and line $y=4$ in first quadrant is

• A. $2$ sq. units
• B. $4$ sq. units
• C. $6$ sq. units
• D. None of these

Answer 1

The correct option is B $4$ sq. units

Let $P=(h,k)$ be a general point in the first quadrant such
$d(P,A)=d(P,O)$
$|h-3|+|k-2|=|h|+|k|=h+k$ .... (i)
[h and k are +ve, point P(h, k) being in first quadrant]
If h < 3, k < 2, then (h, k) lies in region l. Then,
$3-h+2-k=h+k\Rightarrow h+k=5/2$
If h > 3, k < 2, then (h, k) lies in region II. Then,
$h-3+2-k=h+k$
$\Rightarrow k=-1/2$ (not possible)
If h > 3, k > 2 then (h, k) lies in region III. Then,
$h-3+k-2=h+k\Rightarrow -5=0$ (not possible)
If h < 3, k > 2, then (h, k) lies in region IV. Then,
$3-h+k-2=h+k\Rightarrow h=1/2$
Hence, the required set consists of line segment $x+y=5/2$ of finite length as shown in the first region and the ray
$x=1/2$ in the fourth region.
Area of region OBCDEFO
A $=$ area of trapezium OBCF + area of rectangle FCDE
$=\frac{1}{2}\times (\frac{5}{2}+\frac{1}{2})\times 2+\frac{1}{2}\times 2=4$