Four sides of a quadrilateral are given by the equation xy (x-2) (y-3) = 0, then the equation of the line parallel to x – 4y = 0 that divides the quadrilateral into two equal parts is

# Four sides of a quadrilateral are given by the equation xy (x-2) (y-3) = 0, then the equation of the line parallel to x – 4y = 0 that divides the quadrilateral into two equal parts is

1. A

x – 4y + 5 = 0

2. B

x – 4y -5 =0

3. C

x – 4y + 1 = 0

4. D

x – 4y – 1 =0

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### Solution:

Given sides xy ( x- 2) (y – 3) = 0

x = 0, y = 0, x - 2 = 0, y – 3 = 0

Area of rectangle = 2 x 3 = 6

Equation of line parallel to x – 4y = 0 is y = ¼ x + c

This line cuts rectangle at p ( 0, c) and Q ( 2, c+1/2)

Area of trapezium = ½ x (sum of parallel lines) x ( distance b/w parallel lines)

$\begin{array}{l}⇒\text{\hspace{0.17em}}\frac{1}{2}×\text{\hspace{0.17em}}\left(c+c+\frac{1}{2}\right)×2=3\\ ⇒\text{\hspace{0.17em}}2c+\frac{1}{2}=3\\ ⇒\text{\hspace{0.17em}}2c=3-\frac{1}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}c\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{5}{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}The\text{\hspace{0.17em}\hspace{0.17em}}line\text{\hspace{0.17em}\hspace{0.17em}}y=\frac{1}{4}x+\frac{5}{4}\\ ⇒\text{\hspace{0.17em}}x-4y+5=0\end{array}$  +91

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