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

A
x – 4y + 5 = 0
B
x – 4y -5 =0
C
x – 4y + 1 = 0
D
x – 4y – 1 =0

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)

$\begin{array}{l} \Rightarrow \frac{1}{2} \times (c + c + \frac{1}{2}) \times 2 = 3 \\ \Rightarrow 2 c + \frac{1}{2} = 3 \\ \Rightarrow 2 c = 3 - \frac{1}{2} \\ \Rightarrow c = \frac{5}{2} \\ ∴ T h e l i n e y = \frac{1}{4} x + \frac{5}{4} \\ \Rightarrow x - 4 y + 5 = 0 \end{array}$

Solution:

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