Question

The sides of a rectangle are $x=0,y=0,x=4,y=3$. The equation of the straight line having slope $\frac{1}{2}$ that divides the rectangle into two equal halves, is

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

Answer 1

The correct option is A 2y = x + 1

Let the required line be $y=mx+\frac{c}{2}$
Since, given slope $m=\frac{1}{2}$
$\Rightarrow y=\frac{1}{2}m+\frac{c}{2}$
$\therefore 2y=x+c$.
When $x=4$ then $2y=4+c\Rightarrow y=\frac{4+c}{2}$ and
When $x=0$ then $2y=0+c\Rightarrow y=\frac{c}{2}$
Thus, we have, $A_1\equiv (4,\frac{4+c}{2}),B_1\equiv (0,\frac{c}{2})$
Area of rectangle ABCD is $=4\times 3=12$ sq unit
Thus, area of trapezium $OA{A}_1{B}_1$ is half of the area of rectangle $ABCD$
$\Rightarrow \frac{12}{2}=\frac{1}{2}(a+b)h$, where $a=\frac{c}{2}$ and $b=\frac{4+c}{2}$ are lengths of parallel sides and $h=4$ is distance between parallel sides.
$\Rightarrow 6=\frac{1}{2}\left(4\right)(\frac{c}{2}+\frac{4+c}{2})\Rightarrow c=1$
Thus required line is $2y=x+1$