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−c1)(d2−c2)a1b2−a2b1∣∣ sq. units.
Deduce the condition for these lines to form a rhombus.
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=∣∣m1−m21+m1m2∣∣
=−a2b2+a1b11−a1a2b1b2
tan θ=a2b1−a1b2a1a2+b1a2
⇒sin θ=a2b1−a1b2√(a21+b21)+(a22+b22)
p1= Distance between AB and AD.
=∣∣ ∣∣c1−d1√a21+b21∣∣ ∣∣
p2= Distance between AD and BC.
=∣∣ ∣∣c2−d2√a22+b22∣∣ ∣∣
∴ Area of parallelogram is
=|c1−d1||c2−d2||a2b1−a1b2|
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∣∣(c1−d1)a2b1−a1b2∣∣ or 2∣∣c2−d2a2b1−a1b2∣∣