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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 (d1c1)(d2c2)a1b2a2b1 sq. units.

Deduce the condition for these lines to form a rhombus.

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Solution

Let ABCD be a parallelogram the equation of whose sides AB, BC, CD and DA are a1x1,b1y1,c1=0,a2x+b2y+c2=0,a1x+b1y+d1=0 and a2x+b2y+d2=0

Let p1, and p2 be the distance between the pairs of parallel side of ABCD.

sin θ p1AD=p2AB

AD=p1sin θ and AB=p2sin θ

Area of ABCD =AB×p1=p1p2sin θ

or AD×p2=p1p2sin θ

Now,

m1= slope of AB =a1b1

m2= slope of AD =a2b1

Since θ is angle between AB and AC.

tan θ=m1m21+m1m2

=a2b2+a1b11a1a2b1b2

tan θ=a2b1a1b2a1a2+b1a2

sin θ=a2b1a1b2(a21+b21)+(a22+b22)

p1= Distance between AB and AD.

=∣ ∣c1d1a21+b21∣ ∣

p2= Distance between AD and BC.

=∣ ∣c2d2a22+b22∣ ∣

Area of parallelogram is

=|c1d1||c2d2||a2b1a1b2|

Hence proved.

(ii) Rhombus is a parallelogram with all side equal

p1=p2

Modifying the formula of area of parallelogram divided above.

The area of rhombus

=p1p2sin θ

=2p1sin θ=2p2sin θ

=2(c1d1)a2b1a1b2 or 2c2d2a2b1a1b2


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