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Combination with Restrictions

Find the area...

Question

Find 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 $3 x - 2 y - 2 a = 0$

A

\frac{2 a^{2}}{13}

square units

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B

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

square units

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C

\frac{2 a}{5}

square units

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D

\frac{4 a^{2}}{5}

square units

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Solution

The correct option is B $\frac{2 a^{2}}{5}$ square units
Given lines
$2 x - 3 y + a = 0$ and $2 x - 3 y + 3 a = 0$
On Comparing both equation with $y = m x + c$
$m_{1} = \frac{2}{3}, c_{1} = \frac{a}{3}, c_{2} = a$ and
$3 x - 2 y - a = 0$ and $3 x - 2 y - 2 a = 0$
On Comparing both equation with $y = m x + d$
$m_{2} = \frac{3}{2}, d_{1} = \frac{- a}{2}, d_{2} = - a$

Area $= ∣ ∣ ∣ \frac{(c_{1} - c_{2}) (d_{1} - d_{2})}{m_{1} - m_{2}} ∣ ∣ ∣$

Area $= ∣ ∣ ∣ ∣ ∣ ∣ \frac{(\frac{a}{3} - a) (\frac{- a}{2} + a)}{\frac{2}{3} - \frac{3}{2}} ∣ ∣ ∣ ∣ ∣ ∣$

Area $= ∣ ∣ ∣ ∣ ∣ ∣ \frac{\frac{- 2 a}{3} (\frac{a}{2})}{\frac{- 5}{6}} ∣ ∣ ∣ ∣ ∣ ∣$

Area $= ∣ ∣ ∣ ∣ ∣ ∣ \frac{\frac{- 2 a^{2}}{6}}{\frac{- 5}{6}} ∣ ∣ ∣ ∣ ∣ ∣$

Area $= \frac{2 a^{2}}{5}$

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Similar questions

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

3 x - 2 y - 2 a = 0

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

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

3 x - 4 y + 2 a = 0

Q. Prove that the area of the parallelogram formed by the lines 3x − 4y + a = 0, 3x − 4y + 3a = 0, 4x − 3y − a = 0 and 4x − 3y − 2a = 0 is

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

Q. The area of the parallelogram formed by the lines

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

,

2 x - 3 y + 3 a = 0

3 x - 2 y = 2 a

Q. if the lines

3 x - 4 y - 7 = 0

2 x - 3 y - 5 = 0

are two diameter of a circle of area

49 π

square units the equation of the circle is

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