Question

The four sides of a quadrilateral are given by the equation $xy(x-2)(y-3)=0.$ The equation of the line parallel to $x-4y=0$ that divides the quadrilateral in two equal areas is

• A. $x-4y+5=0$
• B. $x-4y-5=0$
• C. $4y=x+1$
• D. $4y+1=x$

Answer 1

The correct option is A $x-4y+5=0$

The four sides of a quadrilateral are given by the equation $xy(x-2)(y-3)=0$.

If we separate them, The lines will be, $x=0$, $y=0$, $x=2$ and $y=3$

The four sides create a rectangle with sides $2$ units and $3$ units.

The area of Rectangle is $6$ square units.

Now if some line parallel to line $x-4y=0$ cuts the rectangle into two equal areas,

The slope of the lines will be the same as the slope of the line $x-4y=0$, which is $\frac{1}{4}$

Let's suppose the line is $y=mx+c$, which cuts the rectangle into two equal areas. The value of $m$ will be $\frac{1}{4}$

$\Rightarrow y=\frac{1}{4}x+c$

From The figure you can see that this lines cuts the rectangle at two points, $P(0,c)$ and $Q(2,c+\frac{1}{2})$

The lines cut the rectangle in two equal area trapeziums.

We know that the area of trapezium $=\frac{1}{2}\times \left(\text{sum of parallel lines}\right)\times \text{Distance between parallel lines}$

$\Rightarrow \text{Area of one trapezium = 6 square units}=\frac{1}{2}\times (c+c+\frac{1}{2})\times \left(2\right)=6$

$\Rightarrow 2c=6-\frac{1}{2}$

$\Rightarrow c=\frac{5}{4}$

Hence The equation of the cutting the given Quadrilateral in two equal area pieces is given by $y=\frac{1}{4}x+\frac{5}{4}$

$\Rightarrow x-4y+5=0$

Hence the correct option is $A$