The area of a parallelogram formed by the lines ax- by- c-0- is-

The correct option is B 2c^2/ab ax± by± c=0⇒x/± c/a+y/± c/b=1 which meets on axes at A(c/a, 0), C( c/a, 0), B( 0, c/b) and D( 0, c/b) The ...

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Standard VIII

Mathematics

Diagonals of a Rhombus Bisect Each-Other at Right Angles

The area of a...

Question

The area of a parallelogram formed by the lines $a x \pm b y \pm c = 0$ , is?

\frac{c^{2}}{a b}

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\frac{2 c^{2}}{a b}

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\frac{c^{2}}{2 a b}

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None of these

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Solution

The correct option is B $\frac{2 c^{2}}{a b}$
$a x \pm b y \pm c = 0 \Rightarrow \frac{x}{\pm c / a} + \frac{y}{\pm c / b} = 1$
which meets on axes at $A (\frac{c}{a}, 0), C (\frac{- c}{a}, 0), B (0, \frac{c}{b})$ and $D (0, \frac{- c}{b})$
Therefore, the diagonals AC and BD of quadrilateral ABCD are perpendicular, hence it is a rhombus whose area is given by $= \frac{1}{2} A C \times B D = \frac{1}{2} \times \frac{2 c}{a} \times 2 \frac{c}{b} = \frac{2 c^{2}}{a b}$ .

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Similar questions

Q. Show that the four lines

a x \pm b y \pm c = 0

enclose a rhombus whose area is

\frac{2 c^{2}}{a b}

Prove that the area of the parallelogram formed by the lines.

$a_{1} x + b_{1} y + c_{1} = 0, a_{1} x + b_{1} y + d_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0, a_{2} x + b_{2} y + d_{2} = 0$ is $∣ ∣ \frac{(d_{1} - c_{1}) (d_{2} - c_{2})}{a_{1} b_{2} - a_{2} b_{1}} ∣ ∣$ sq. units.

Deduce the condition for these lines to form a rhombus.

Q. Prove that the area of the parallelogram formed by the lines a₁x + b₁y + c₁ = 0, a₁x + b₁y + d₁ = 0, a₂x + b₂y + c₂ = 0, a₂x + b₂y + d₂ = 0 is

|\frac{(d_{1} - c_{1}) (d_{2} - c_{2})}{a_{1} b_{2} - a_{2} b_{1}}|

sq. units.
Deduce the condition for these lines to form a rhombus.

Q. Assertion :The two straight lines given by the equation

(a^{2} - 3 b^{2}) x^{2} + 8 a b x y + (b^{2} - 3 a^{2}) y^{2} = 0

from with the line

a x + b y + c = 0

a triangle of area

\frac{c^{2}}{\sqrt{3} (a^{2} + b^{2})}

Reason: The triangles formed by the lines

(a^{2} - 3 b^{2}) x^{2} + 8 a b x y + (b^{2} - 3 a^{2}) y^{2} = 0

and

a x + b y + c = 0

is an equilateral triangle.

Q. Show that the straight lines

(A^{2} - 3 B^{2}) x^{2} + 8 A B x y + (B^{2} - 3 A^{2}) y^{2} = 0

form with the line

A x + B y + C = 0

an equilateral triangle of area

\frac{C^{2}}{\sqrt 3. (A^{2} + B^{2})}

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The area of a parallelogram formed by the lines ax±by±c=0, is?

The correct option is B 2c2abax±by±c=0⇒x±c/a+y±c/b=1which meets on axes at A(ca,0),C(−ca,0),B(0,cb) and D(0,−cb)Therefore, the diagonals AC and BD of quadrilateral ABCD are perpendicular, hence it is a rhombus whose area is given by =12AC×BD=12×2ca×2cb=2c2ab.

The area of a parallelogram formed by the lines $a x \pm b y \pm c = 0$ , is?