Prove that the area of the parallelogram formed by the lines a1 x-b1 y-c1-0- a1 x-b1 y-d1-0- a2 x-b2 y-c2-0- a2 x-b2 y-d2-0 is d1-c1d2-c2a1b2-a2b1 sq- units- Deduce the condition for these lines to form a rhombus-

The given lines are a1x + b1y + c1 = 0 nbsp; nbsp; nbsp; nbsp; nbsp; ... (1) a1x + b1y + d1 = 0 nbsp; nbsp; nbsp; nbsp; nbsp; ... (2) a2x ...

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Byju's Answer

Standard IX

Mathematics

Parallelograms on the Same Base and between the Same Parallels

Prove that th...

Question

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.

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Solution

The given lines are

a₁x + b₁y + c₁ = 0 ... (1)

a₁x + b₁y + d₁ = 0 ... (2)

a₂x + b₂y + c₂ = 0 ... (3)

a₂x + b₂y + d₂ = 0 ... (4)

The area of the parallelogram formed by the lines a₁x + b₁y + c₁ = 0, a₁x + b₁y + d₁ = 0, a₂x + b₂y + c₂ = 0 and a₂x + b₂y + d₂ = 0 is given below:

$Area = |\frac{(c_{1} - d_{1}) (c_{2} - d_{2})}{|\begin{matrix} a_{1} & a_{2} \\ b_{1} & b_{2} \end{matrix}|}|$

$∵ |\begin{matrix} a_{1} & a_{2} \\ b_{1} & b_{2} \end{matrix}| = a_{1} b_{2} - a_{2} b_{1}$

$∴ Area = |\frac{(c_{1} - d_{1}) (c_{2} - d_{2})}{a_{1} b_{2} - a_{2} b_{1}}| = |\frac{(d_{1} - c_{1}) (d_{2} - c_{2})}{a_{1} b_{2} - a_{2} b_{1}}|$

If the given parallelogram is a rhombus, then the distance between the pair of parallel lines are equal.

$∴ |\frac{c_{1} - d_{1}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2}}}| = |\frac{c_{2} - d_{2}}{\sqrt{{a_{2}}^{2} + {b_{2}}^{2}}}|$

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Similar questions

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, the equations of whose
sides are

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

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

Q. Equation/equations of the planes bisecting the angles between the planes

a_{1} x + b_{1} y + c_{1} z + d_{1} = 0

and

a_{2} x + b_{2} y + c_{2} z + d_{2} = 0

is given by -

The condition for three lines $a_{1}$ x+ $b_{1}$ y+ $c_{1}$ = 0, $a_{2}$ x+ $b_{2}$ y+ $c_{2}$ = 0 and $a_{3}$ x+ $b_{3}$ y+ $c_{3}$ = 0 , to be concurrent is

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

(A);

(a_{1} x + b_{1} y + c_{1}) + (a_{2} x + b_{2} y + c_{2}) + a_{3} x + b_{3} y + c_{3}) = 0,

then the lines

a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0, a_{3} x + b_{3} y + c_{3} = 0

can not be parallel.
(R): If sum of three straight lines is identically

0,

then they are either concurrent or parallel.

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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 d1-c1d2-c2a1b2-a2b1 sq. units. Deduce the condition for these lines to form a rhombus.