Prove that the area of the parallelogram, the equations of whose
sides are
$a_{1} x + b_{1} y + c_{1} = 0, a_{1} x + b_{1} y + d_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0, a_{2} x + b_{2} y + d_{2} = 0$
is $\frac{(d_{1} - c_{1}) (d_{2} - c_{2})}{a_{1} b_{2} - a_{2} b_{1}}$

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Solution

$p_{1} = \frac{d_{1} - c_{1}}{\sqrt (a_{1}^{2} + b_{1}^{2})}, p_{2} = \frac{d_{2} - c_{2}}{\sqrt (a_{2}^{2} + b_{2}^{2})}$
Also $sin θ = \frac{a_{1} b_{2} - a_{2} b_{1}}{\sqrt (a_{1}^{2} + b_{1}^{2}) \sqrt (a_{2}^{2} + b_{2}^{2})}$
Area of parallelogram
$\frac{p_{1} - p_{2}}{sin θ} = \frac{(c_{1} - d_{2}) (c_{1} - d_{2})}{a_{1} b_{2} - a_{2} b_{1}}$
Note: If these lines enclose a rhombus, then
$p_{1} = p_{2}$
$∴ (c_{1} - d_{2}) (a_{2}^{2} + b_{2}^{2}) = (c_{2} - d_{2}) (a_{1}^{2} + b_{1}^{2}) .$

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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 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. 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 -

Q. The two lines a₁x + b₁y = c₁ and a₂x + b₂y = c₂ are perpendicular if
(a) a₁a₂ + b₁b₂ = 0
(b) a₁b₂ + a₂b₁
(c) a₁b₁ + a₂b₂ = 0
(d) a₁b₂ + a₂b₁ = 0

For the pair of linear equations $a_{1} x$ + $b_{1} y$ + $c_{1} = 0$ and $a_{2} x$ + $b_{2} y$ + $c_{2} = 0$ , the solution set are $x = \frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ and $y = \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ . They can also be written as:

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Prove that the area of the parallelogram, the equations of whose sides area1x+b1y+c1=0,a1x+b1y+d1=0a2x+b2y+c2=0,a2x+b2y+d2=0is (d1−c1)(d2−c2)a1b2−a2b1

p1=d1−c1√(a21+b21),p2=d2−c2√(a22+b22)Also sinθ=a1b2−a2b1√(a21+b21)√(a22+b22)Area of parallelogramp1−p2sinθ=(c1−d2)(c1−d2)a1b2−a2b1Note: If these lines enclose a rhombus, then p1=p2∴(c1−d2)(a22+b22)=(c2−d2)(a21+b21).

Prove that the area of the parallelogram, the equations of whose
sides are
$a_{1} x + b_{1} y + c_{1} = 0, a_{1} x + b_{1} y + d_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0, a_{2} x + b_{2} y + d_{2} = 0$
is $\frac{(d_{1} - c_{1}) (d_{2} - c_{2})}{a_{1} b_{2} - a_{2} b_{1}}$