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The four sides of a quadrilateral are given by the equation $xy\left( x-2 \right)\left( y-3 \right)=0$. 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). $4y+1=x$

Last updated date: 23rd Feb 2024
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IVSAT 2024
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Hint: First try to draw the quadrilateral by using the given equation. Then calculate the area of the quadrilateral. Find out the intersecting points of the parallel line and the quadrilateral. Then form an equation using the given condition.

Complete step-by-step answer:

The given equation of the quadrilateral is, $xy\left( x-2 \right)\left( y-3 \right)=0$.
The equations of the four sides of this given quadrilateral are:
  & x=0,y=0 \\
 & x-2=0\Rightarrow x=2 \\
 & y-3=0\Rightarrow y=3 \\
These are the four sides. So let us draw the quadrilateral,

In the above picture $ABCD$ is the quadrilateral. Where,
  & AB:x=0 \\
 & BC:x=2 \\
 & CD:y=3 \\
 & DA:y=0 \\
Coordinates of the four points of the quadrilateral are,
  & A\left( 0,0 \right) \\
 & B\left( 2,0 \right) \\
 & C\left( 2,3 \right) \\
 & D\left( 0,3 \right) \\
We can clearly see that this is a rectangle. So the area of $ABCD$ is,
$\left( 2\times 3 \right)=6$ square unit.
We have to divide this area in two equal parts by a line which is parallel to $x-4y=0$.
We know that the equation of the line which is parallel to the above line is:
$x-4y+k=0.......(a)$, where $k$ is a constant.
We can write it as:
  & \Rightarrow x-4y=-k \\
 & \Rightarrow \dfrac{x}{-k}+\dfrac{y}{\dfrac{k}{4}}=1 \\
Therefore the line (a) cuts the x axis at $E\left( -k,0 \right)$ and y axis at $F\left( 0,\dfrac{k}{4} \right)$.
Line (a) cuts the side $BC$ at the point $G$. Therefore,
$2-4y=-k$, by putting the value $x=2$ in equation (a)
  & \Rightarrow -4y=-k-2 \\
 & \Rightarrow y=\dfrac{k+2}{4} \\
Hence, the coordinate of $G$ is $\left( 2,\dfrac{k+2}{4} \right)$.
Now, we have to find out the area of $FGCD$.
Let us draw a perpendicular from the point $F$ on the side $BC$, which meets the line $BC$ at $H$.
Now the area of $FGCD$ = The area of rectangle $FHCD$ - The area of the triangle $FHG$.
We can see that:
$FH=AB=DC=2$ unit.
$FD=AD-AF=3-\dfrac{k}{4}$ unit.
Area of $FHCD$ is $2\times \left( 3-\dfrac{k}{4} \right)=6-\dfrac{k}{2}$ square units.
$GH=BG-BH=BG-AF=\dfrac{k+2}{4}-\dfrac{k}{4}=\dfrac{k+2-k}{4}=\dfrac{2}{4}=\dfrac{1}{2}$ unit.
Therefore are of the triangle $FHG$ is:
$\dfrac{1}{2}\times FH\times GH=\dfrac{1}{2}\times 2\times \dfrac{1}{2}=\dfrac{1}{2}$ square unit.
Hence, the area of $FGCD$ is $=\left( 6-\dfrac{k}{2} \right)-\dfrac{1}{2}$ square unit, which is half of the area of the quadrilateral.
  & \left( 6-\dfrac{k}{2} \right)-\dfrac{1}{2}=\dfrac{6}{2} \\
 & \Rightarrow 6-\dfrac{k}{2}=3+\dfrac{1}{2} \\
 & \Rightarrow \dfrac{k}{2}=6-\dfrac{7}{2} \\
 & \Rightarrow \dfrac{k}{2}=\dfrac{12-7}{2} \\
 & \Rightarrow k=5 \\
Hence, the value of k is 5.
The equation of the straight line is:
Therefore option (a) is correct.

Note: Alternatively we can find out the answer by cross checking the options. Take the lines one by one. At first we have to find out the coordinates of the intersecting points of the line and the quadrilateral. Then we have to find out the area and check if it is 3 square units or not.
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