GATE CS 2025 Set 2 — Question 9

Let the vertices of the square PQRS be represented in a Cartesian coordinate system. Let $P$ be at $(0, 2)$, $Q$ at $(0, 0)$, $R$ at $(2, 0)$, and $S$ at $(2, 2)$. The side length of the square is 2 cm.The point $L$ lies on the side $QR$, which is on the x-axis. Let the coordinates of $L$ be $(x, 0)$, where $0 \le x \le 2$.The rectangle is PLMN. The side $PL$ connects $P(0, 2)$ and $L(x, 0)$.
The length of side $PL$ is the distance between these two points:$$PL = \sqrt{(x - 0)^2 + (0 - 2)^2} = \sqrt{x^2 + 4}$$The side $MN$ is parallel to the side $PL$. The width of the rectangle, $PN$, is the perpendicular distance between the line segment $PL$ and the line segment $MN$.
Since the side $MN$ passes through the corner $S(2, 2)$, the width $PN$ is equal to the perpendicular distance from point $S(2, 2)$ to the line passing through $P$ and $L$.The equation of the line passing through $P(0, 2)$ and $L(x, 0)$ can be found using the intercept form $\frac{X}{a} + \frac{Y}{b} = 1$:$$\frac{X}{x} + \frac{Y}{2} = 1 \implies 2X + xY - 2x = 0$$The perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Here, point $S$ is $(2, 2)$, and the line is $2X + xY - 2x = 0$.$$PN = d = \frac{|2(2) + x(2) - 2x|}{\sqrt{2^2 + x^2}} = \frac{|4 + 2x - 2x|}{\sqrt{4 + x^2}} = \frac{4}{\sqrt{4 + x^2}}$$The area of the rectangle PLMN is the product of its length and width:$$\text{Area} = PL \times PN = \sqrt{x^2 + 4} \times \frac{4}{\sqrt{x^2 + 4}} = 4$$Thus, the area of the rectangle is 4 cm$^2$.