Area of the quadrilateral formed by the lines $4 y - 3 x - a = 0, 3 y - 4 x + a = 0, 4 y - 3 x - 3 a = 0$ , $3 y - 4 x + 2 a = 0$ in sq.units is

\frac{a^{2}}{5}

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\frac{a^{2}}{7}

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\frac{2 a^{2}}{7}

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\frac{2 a^{2}}{9}

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Solution

The correct option is D $\frac{2 a^{2}}{7}$
We have,
$4 x - 3 x - a = 0$ ---(1)
$3 y - 4 x + a = 0$ ---(2)
$4 y - 3 x - 3 a = 0$ ---(3)
$3 y - 4 x + 2 a = 0$ ---(4)
$4 y - 3 a - a = 0$ and $3 y - 4 x + a = 0$
$4 x - 3 x - 3 a = 0$ $3 y - 4 x + 2 a = 0$
So,
$m_{1} = \frac{3}{4}$ $m_{2} = \frac{4}{3}$
Perpendicular distance perpendicular distance between these lines
Between this lines $= ∣ ∣ ∣ \frac{2 a}{5} ∣ ∣ ∣$ $= ∣ ∣ ∣ \frac{a}{5} ∣ ∣ ∣$
From 1 & 2, From 1& 4,
$y = a, x = a$ $y = \frac{10}{7} a, x = \frac{11}{7} a$
$A B = \frac{5 a}{7}$
So, area of parallelogram formed is $h \times b$
$= \frac{5 a}{7} \times \frac{2 a}{5} = \frac{2 a^{2}}{7}$

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Similar questions

Q. The area of the quadrilateral formed by the lines

4 x - 3 y - a = 0, 3 x - 4 y + a = 0

4 x - 3 y - 3 a = 0

and

3 x - 4 y + 2 a = 0

Q. The area of the quadrilateral formed by the lines

4 x - 3 y - a = 0, 3 x - 4 y + a = 0

4 x - 3 y - 3 a = 0

and

3 x - 4 y + 2 a = 0

Q. For the quadrilateral formed by the lines

4 y - 3 x - 1 = 0, 3 y + 4 x + 1 = 0, 4 y - 3 x - 2 = 0

and

3 y + 4 x + 2 = 0

, which among the following statements are true

Prove that the area of the parallelogram formed by the lines $3 x - 4 y + a = 0$ , $3 x - 4 y + 3 a = 0$ , $4 x - 3 y - a = 0$ and $4 x - 3 y - 2 a = 0$ is $\frac{2 a^{2}}{7}$ sq. units.

Q. Show that the area of the parallelogram formed by the lines

2 x - 3 y + a = 0

3 x - 2 y - a = 0, 2 x - 3 y + 3 a = 0

and

3 x - 2 y - 2 a = 0

\frac{2 a^{2}}{5}

sq. units.

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Area of the quadrilateral formed by the lines 4y−3x−a=0,3y−4x+a=0,4y−3x−3a=0, 3y−4x+2a=0 in sq.units is

Area of the quadrilateral formed by the lines $4 y - 3 x - a = 0, 3 y - 4 x + a = 0, 4 y - 3 x - 3 a = 0$ , $3 y - 4 x + 2 a = 0$ in sq.units is